Khôi Bùi

Nhìn nhầm 89000 với 8900

Sửa lại

a ) ...

Khối lượng của HHCN là :

m = D . V = 8900 . 0 , 01 = 89 ( kg )

Trọng lượng hay áp lực của HHCN là :

P = F = 10 . m = 10 . 89 = 890 ( N )

ÁP suất của vật khi vật có s mặt đáy nhỏ nhất là :

p = F/s = 890 / 0 ,02 = 44500 ( N/m^2 )

b ) ...

Áp suất của vật khi vật có s mặt đáy LN là :

p = F/s = 890/0,1 = 8900 ( N/m^2 )

Đ/s : ...

a ) Do hình hộp chữ nhật có kích thước 50 . 20 . 10 ( cm )

Để s mặt đáy nhỏ nhất thì chiều cao LN

=> Chiều cao là 50 cm , s mặt đáy là : 20 . 10 = 200 ( cm^2 )

Đổi : 200 cm^2 = 0 , 02 m^2

Thể tích của HHCN là : 50 . 20 . 10 = 10000 ( cm^3 ) = 0 ,01 ( m^3 )

Khối lượng của HHCN là :

m = D . V = 89000 . 0 , 01 = 890 ( kg )

Trọng lượng hay áp lực của HHCN là :

P = F = 10 . m = 890 . 10 = 8900 ( N )

Áp suất của vật khi vật có s mặt đáy nhỏ nhất là :

p = F/s = 8900/0,02 = 445000 ( N/m^2)

b )

Do hình hộp chữ nhật có kích thước 50 . 20 . 10 ( cm )

Để s mặt đáy nhỏ nhất thì chiều cao NN

=> Chiều cao là 10 cm , s mặt đáy là : 20 . 50 = 1000 ( cm^2 )

= 0 , 1 m^2

Áp suất của vật khi vật có s mặt đáy lớn nhất là :

p = F/s = 8900/0,1 = 89000 ( N/m^2)

Đ/s : a ) 445000 N/m^2

b ) 89000 N/m^2

\(\left(x^2+x\right)^2+\left(x^2+x\right)-6=0\)

\(\Leftrightarrow\left(x^2+x\right)^2-2\left(x^2+x\right)+3\left(x^2+x\right)-6=0\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)+3\left(x^2+x-2\right)=0\)

\(\Leftrightarrow\left(x^2+x+3\right)\left(x^2+x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+3=0\\x^2+x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+\dfrac{1}{4}+\dfrac{11}{4}=0\\\left(x+2\right)\left(x-1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2=-\dfrac{11}{4}\left(L\right)\\\left(x+2\right)\left(x-1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

Ta có :

\(x^3+ax^2+2x+b\)

\(=x^3+x^2+x+\left(a-1\right)x^2+x+b\)

\(=x\left(x^2+x+1\right)+\left(a-1\right)x^2+ax-x+a-1+2x-ax-a+b+1\)

\(=x\left(x^2+x+1\right)+\left(a-1\right)x^2+x\left(a-1\right)+a-1+2x-ax-a+b+1\)

\(=x\left(x^2+x+1\right)+\left(a-1\right)\left(x^2+x+1\right)+2x-ax-a+b+1\)

\(=\left(a-1+x\right)\left(x^2+x+1\right)+\left(2-a\right)x+b-a+1\)

\(\Rightarrow\) \(x^3+ax^2+2x+b:\left(x^2+x+1\right)\) dư \(\left(2-a\right)x+b-a+1\)

Để chia hết thì \(\left(2-a\right)x+b-a+1=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2-a=0\\b-a+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=a-1\end{matrix}\right.\)

\(\Leftrightarrow a=2;b=1\)

Vậy ..

a ) \(x\left(x^2-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm2\end{matrix}\right.\)

Vậy ...

b ) \(x^2-5x+6=0\)

\(\Leftrightarrow x^2-2x-3x+6=0\)

\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)

Vậy ...

c ) \(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow x^2+4x+4-\left(x^2-4\right)=0\)

\(\Leftrightarrow x^2+4x+4-x^2+4=0\)

\(\Leftrightarrow4x+8=0\)

\(\Leftrightarrow4x=-8\)

\(\Leftrightarrow x=-2\)

Vậy ...

d ) \(2x^2-x-6=0\)

\(\Leftrightarrow2x^2-4x+3x-6=0\)

\(\Leftrightarrow2x\left(x-2\right)+3\left(x-2\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-3\\x=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=2\end{matrix}\right.\)

Vậy ...

a ) \(x^2-4+\left(x-2\right)^2\)

\(=x^2-4+x^2-4x+4\)

\(=2x^2-4x\)

\(=2x\left(x-2\right)\)

b ) \(x^3+2x^2+x+2\)

\(=x^2\left(x+2\right)+x+2\)

\(=\left(x^2+1\right)\left(x+2\right)\)

c ) \(x^3-2x^2+x-xy^2\)

\(=x\left(x^2-2x+1-y^2\right)\)

\(=x\left[\left(x-1\right)^2-y^2\right]\)

\(=x\left(x-1-y\right)\left(x-1+y\right)\)

d ) \(x^2-3x+2\)

\(=x^2-2x-x+2\)

\(=x\left(x-2\right)-\left(x-2\right)\)

\(=\left(x-1\right)\left(x-2\right)\)

\(\left(a+b\right)\left(a^2-b^2\right)+\left(b+c\right)\left(b^2-c^2\right)+\left(a+c\right)\left(c^2-a^2\right)\)

\(=\left(a+b\right)\left(a^2-b^2\right)-\left(b+c\right)\left(c^2-b^2\right)+\left(a+c\right)\left(c^2-a^2\right)\)

\(=\left(a+b\right)\left(a^2-b^2\right)-\left(b+c\right)\left(a^2-b^2+c^2-a^2\right)+\left(a+c\right)\left(c^2-a^2\right)\)

\(=\left(a+b\right)\left(a^2-b^2\right)-\left(b+c\right)\left(a^2-b^2\right)-\left(b+c\right)\left(c^2-a^2\right)+\left(a+c\right)\left(c^2-a^2\right)\)

\(=\left(a+b-b-c\right)\left(a^2-b^2\right)+\left(c^2-a^2\right)\left(a+c-b-c\right)\)

\(=\left(a-c\right)\left(a-b\right)\left(a+b\right)-\left(a-c\right)\left(a+c\right)\left(a-b\right)\)

\(=\left(a-c\right)\left(a-b\right)\left(a+b-a-c\right)\)

\(=\left(a-c\right)\left(a-b\right)\left(b-c\right)\)

Bài 1

\(x^3-4x^2+8x-8=\left(x^3-8\right)-4x\left(x-2\right)=\left(x-2\right)\left(x^2+2x+4\right)-4x\left(x-2\right)\)

\(=\left(x-2\right)\left(x^2+2x+4-4x\right)=\left(x-2\right)\left(x^2-2x+4\right)\)

Bài 2

\(a^2+b^2-a^2b^2+ab-a-b\)

\(=\left(a^2-a^2b^2\right)+\left(b^2-b\right)+\left(ab-a\right)\)

\(=a^2\left(1-b^2\right)+b\left(b-1\right)+a\left(b-1\right)\)

\(=a^2\left(1-b\right)\left(1+b\right)+\left(a+b\right)\left(b-1\right)\)

\(=-a^2\left(b-1\right)\left(b+1\right)+\left(a+b\right)\left(b-1\right)\)

\(=\left(b-1\right)\left[a+b-a^2\left(b+1\right)\right]\)

\(=\left(b-1\right)\left(a+b-a^2b-a^2\right)\)

\(=\left(b-1\right)\left[\left(a-a^2\right)+b\left(1-a^2\right)\right]\)

\(=\left(b-1\right)\left[a\left(1-a\right)+b\left(1-a\right)\left(1+a\right)\right]\)

\(=\left(b-1\right)\left(1-a\right)\left[a+b\left(1+a\right)\right]\)

\(=\left(b-1\right)\left(1-a\right)\left(a+b+ab\right)\)

Bài 3

\(3x^2+6xy+3y^2-3z^2=3\left(x^2+2xy+y^2-z^2\right)=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y-z\right)\left(x+y+z\right)\)

Bài 4

\(5x\left(x-2\right)-3x^2\left(x-2\right)=x\left(x-2\right)\left(5-3x\right)\)

\(3x\left(x-5y\right)-2y\left(5y-x\right)=3x\left(x-5y\right)+2y\left(x-5y\right)=\left(3x+2y\right)\left(x-5y\right)\)

c ) \(C=\left(x^2-3x+1\right)\left(x^2-3x+1\right)=\left(x^2-3x+1\right)^2\ge0\forall x\)

Dấu " = " xảy ra \(\Leftrightarrow x^2-3x+1=0\)

\(\Leftrightarrow x^2-3x+\dfrac{9}{4}-\dfrac{5}{4}=0\)

\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=\dfrac{5}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{\sqrt{5}}{2}\\x-\dfrac{3}{2}=\dfrac{-\sqrt{5}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{5}+3}{2}\\x=\dfrac{3-\sqrt{5}}{2}\end{matrix}\right.\)

Vậy Min C là : \(0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{5}+3}{2}\\x=\dfrac{3-\sqrt{5}}{2}\end{matrix}\right.\)

d ) \(D=\left(x^2-4x+1\right)\left(x^2-4x+5\right)\)

\(=\left(x^2-4x+3-2\right)\left(x^2-4x+3+2\right)\)

\(=\left(x^2-4x+3\right)^2-4\ge-4\forall x\)

Dấu " = " xảy ra \(\Leftrightarrow x^2-4x+3=0\)

\(\Leftrightarrow x^2-3x-x+3=0\)

\(\Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

Vậy Min D là : \(-4\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

\(\left(x^2-2\right)\left(x^2+x-1\right)-x\left(x^3+x^2-3x-2\right)\)

\(=x^4-2x^2+x^3-2x-x^2+2-x^4-x^3+3x^2+2x\)

\(=\left(x^4-x^4\right)+\left(x^3-x^3\right)+\left(3x^2-2x^2-x^2\right)+\left(2x-2x\right)+2\)

\(=2\)

\(\Rightarrow\) b/t trên ko phụ thuộc vào biến \(\left(đpcm\right)\)

Ta có hình vẽ :

a ) Ta có : MD // AC ( GT ) ; DN // AB ( GT )

hay MD // AN ; DN // AM

=> AMDN là HBH ( đpcm )

b ) Do AMDN là HBH ( cmt )

=> AG = GD

Xét tam giác ADC có :

AG = GD ( cmt ) ; DH = HC ( GT )

=> GH là đường trung bình của tam giác ADC

=> GH // AC

Mà DM // AC ( GT )

=> DM // GH ( đpcm )

P/s : Hình bạn vẽ sai

1 ) \(x^3-2x^2y+xy^2-4x\)

\(=x\left(x^2-2xy+y^2-4\right)\)

\(=x\left[\left(x-y\right)^2-4\right]\)

\(=x\left(x-y-2\right)\left(x-y+2\right)\)

2 ) \(x^4-3x^2+2=x^4-2x^2-x^2+2=x^2\left(x^2-2\right)-\left(x^2-2\right)=\left(x^2-1\right)\left(x^2-2\right)=\left(x-1\right)\left(x+1\right)\left(x^2-2\right)\)

3 ) \(27ay^2-3a-3ab^2+6ab\)

\(=3a\left(9y^2-1-b^2+2b\right)\)

\(=3a\left[9y^2-\left(b^2-2b+1\right)\right]\)

\(=3a\left[9y^2-\left(b-1\right)^2\right]\)

\(=3a\left(3y-b+1\right)\left(3y+b-1\right)\)

4 ) \(y^3-3y^2-3y+1=\left(y^3+1\right)-3y\left(y+1\right)=\left(y+1\right)\left(y^2-y+1\right)-3y\left(y+1\right)\)

\(=\left(y+1\right)\left(y^2-y+1-3y\right)=\left(y+1\right)\left(y^2-4y+1\right)\)

1 ) \(3x^3+15x^2+18x\)

\(=3x\left(x^2+5x+6\right)\)

\(=3x\left(x^2+2x+3x+6\right)\)

\(=3x\left[x\left(x+2\right)+3\left(x+2\right)\right]\)

\(=3x\left(x+3\right)\left(x+2\right)\)

2 ) \(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)

\(=\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left[x^2+2xy+y^2-xz-yz+z^2-3xy\right]\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)

a ) \(\left(3-2\right)\left(9+6+4\right)=\left(3-2\right)\left(3^2+3.2+2^2\right)=3^3-2^3=27-8=19\)

b ) \(64+144+27+108=4^3+3.4^2.3+3.4.3^2+3^3=\left(4+3\right)^3=7^3=343\)

Ta có : \(\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)+2018\)

\(=\left[\left(x-1\right)\left(x-7\right)\right]\left[\left(x-3\right)\left(x-5\right)\right]+2018\)

\(=\left(x^2-x-7x+7\right)\left(x^2-3x-5x+15\right)+2018\)

\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)+2018\)

\(=\left(x^2-8x+11-4\right)\left(x^2-8x+11+4\right)+2018\)

\(=\left(x^2-8x+11\right)^2-16+2018\)

\(=\left(x^2-8x+11\right)^2+2002\ge2002>0\forall x\)

\(\left(đpcm\right)\)

\(A=x\left(x-1\right)\left(x+2\right)\left(x-3\right)-9\)

\(=\left(x^2-x\right)\left(x^2+2x-3x-6\right)-9\)

\(=\left(x^2-x\right)\left(x^2-x-6\right)-9\)

\(=\left(x^2-x-3+3\right)\left(x^2-x-3-3\right)-9\)

\(=\left(x^2-x-3\right)^2-9-9\)

\(=\left(x^2-x-3\right)^2-18\ge-18\forall x\)

Dấu " = " xảy ra \(\Leftrightarrow x^2-x-3=0\)

\(\Leftrightarrow x^2-x+\dfrac{1}{4}-\dfrac{13}{4}=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{13}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{\sqrt{13}}{2}\\x-\dfrac{1}{2}=\dfrac{-\sqrt{13}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{13}+1}{2}\\x=\dfrac{1-\sqrt{13}}{2}\end{matrix}\right.\)

Vậy ...

Xét \(n=2k\) , ta có :

\(\left(2k\right)^2\left[\left(2k\right)^2-1\right]\left[\left(2k\right)^2+2\right]=4k^2\left(2k-1\right)\left(2k+1\right)\left(4k^2+2\right)\)

\(=8k^2\left(2k-1\right)\left(2k+1\right)\left(2k^2+1\right)⋮8\left(1\right)\)

Xét \(n=2k+1\) , ta có :

\(\left(2k+1\right)^2\left[\left(2k+1\right)^2-1\right]\left[\left(2k+1\right)^2+2\right]=\left(2k+1\right)^2.2k\left(2k+2\right)\left(4k^2+4k+1+2\right)\)

\(=\left(2k+1\right)^2.4k\left(k+1\right)\left(4k^2+4k+3\right)⋮8\left(2\right)\)

( do \(k\left(k+1\right)⋮2\Rightarrow4k\left(k+1\right)⋮8\) )

Với n \(⋮3\Rightarrow n^2⋮9\) \(\Rightarrow n^2\left(n^2-1\right)\left(n^2+2\right)⋮9\left(3\right)\)

Với n \(⋮3̸\) \(\Rightarrow n^2:3\) ( dư 1 ) \(\Rightarrow n^2-1⋮3\Rightarrow n^2+2⋮3\)

Do \(n\left(n-1\right)\left(n+1\right)⋮3\Rightarrow n^2\left(n-1\right)\left(n+1\right)\left(n^2+2\right)⋮9\left(4\right)\)

Từ ( 1 ) ; ( 2 ) ; ( 3 ) ; ( 4 )

\(\Rightarrow n^6+n^4-2n^2⋮72\left(đpcm\right)\)

\(n^6+n^4-2n^2\)

\(=n^2\left(n^4+n^2-2\right)\)

\(=n^2\left[\left(n^4-1\right)+n^2-1\right]\)

\(=n^2\left[\left(n^2-1\right)\left(n^2+1\right)+n^2-1\right]\)

\(=n^2\left(n^2-1\right)\left(n^2+1+1\right)\)

\(=n^2\left(n^2-1\right)\left(n^2+2\right)\)

\(=n\left(n-1\right)\left(n+1\right)n\left(n^2+2\right)\)

Do \(xy\ne0\Rightarrow x;y\ne0\)

Ta có : \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)

\(\Leftrightarrow a^2x^2+b^2x^2+a^2y^2+b^2y^2=a^2x^2+2axby+b^2y^2\)

\(\Leftrightarrow b^2x^2+a^2y^2=2axby\)

\(\Leftrightarrow b^2x^2+a^2y^2-2axby=0\)

\(\Leftrightarrow\left(bx-ay\right)^2=0\)

Do \(\left(bx-ay\right)^2\ge0\Rightarrow bx-ay=0\)

\(\Rightarrow bx=ay\)

\(\Rightarrow\dfrac{a}{x}=\dfrac{b}{y}\left(đpcm\right)\)

Vội nên đánh sai :

\(a\left(a+1\right)+3=a^2+a+3=a^2+a+\dfrac{1}{4}+\dfrac{11}{4}=\left(a+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\)~

a ) \(-x^2+4x-5\)

\(=-\left(x^2-4x+4\right)-1\)

\(=-\left(x-2\right)^2-1\le-1< 0\forall x\left(đpcm\right)\)

b ) \(x^4+3x^2+3=x^4+3x^2+\dfrac{9}{4}+\dfrac{3}{4}=\left(x^2+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\left(đpcm\right)\)

c ) \(\left(x^2+2x+3\right)\left(x^2+2x+4\right)+3\)

Đặt \(x^2+2x+3=a\) . Khi đó , ta có :

\(x\left(x+1\right)+3=x^2+x+3=x^2+x+\dfrac{1}{4}+\dfrac{11}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\)

\(=\left(x^2+2x+3+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\)

\(=\left(x^2+2x+\dfrac{7}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}>0\forall x\left(đpcm\right)\)

a ) Đề sai

b ) \(x^2-x+1=x^2-x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\left(đpcm\right)\)

c ) \(x-x^2-2=-\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{7}{4}=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{7}{4}\le-\dfrac{7}{4}< 0\forall x\left(đpcm\right)\)

Đề sai : \(x^3-y^3+z^3+3xyz\)

\(=\left(x^3-3x^2y+3xy^2-y^3\right)+z^3+3xyz+3x^2y-3xy^2\)

\(=\left(x-y\right)^3+z^3+3xy\left(z+x-y\right)\)

\(=\left(x-y+z\right)\left[\left(x-y\right)^2-\left(x-y\right)z+z^2\right]+3xy\left(x-y+z\right)\)

\(=\left(x-y+z\right)\left[x^2-2xy+y^2-xz+yz+z^2+3xy\right]\)

\(=\left(x-y+z\right)\left(x^2+y^2+z^2+xy-yz-xz\right)\)

Ta có : \(x^3+2x^2+3x+2=y^3\)

\(\Leftrightarrow y^3-x^3=2x^2+3x+2\)

Do \(2x^2+3x+2\)

\(=2\left(x^2+\dfrac{3}{2}x+1\right)\)

\(=2\left(x^2+\dfrac{3}{2}x+\dfrac{9}{16}+\dfrac{7}{16}\right)\)

\(=2\left[\left(x+\dfrac{3}{4}\right)^2+\dfrac{7}{16}\right]\)

\(=2\left(x+\dfrac{3}{4}\right)^2+\dfrac{7}{8}\ge\dfrac{7}{8}>0\forall x\)

\(\Rightarrow y^3-x^3>0\)

\(\Rightarrow y^3>x^3\left(1\right)\)

Lại có : \(\left(x+2\right)^3=x^3+6x^2+12x+8=x^3+2x^2+3x+2+4x^2+9x+6\)

\(=y^3+4x^2+9x+6\)

\(\Rightarrow\left(x+2\right)^3-y^3=4x^2+9x+6\)

Do \(4x^2+9x+6\)

\(=4\left(x^2+\dfrac{9}{4}x+\dfrac{3}{2}\right)\)

\(=4\left(x^2+\dfrac{9}{4}x+\dfrac{81}{72}+\dfrac{3}{8}\right)\)

\(=4\left[\left(x+\dfrac{9}{8}\right)^2+\dfrac{3}{8}\right]\)

\(=4\left(x+\dfrac{9}{8}\right)^2+\dfrac{3}{2}\ge\dfrac{3}{2}>0\forall x\)

\(\Rightarrow\left(x+2\right)^3>y^3\left(2\right)\)

Từ ( 1 ) ; ( 2 )

\(\Rightarrow x^3< y^3< \left(x+2\right)^3\)

\(\Rightarrow y^3=\left(x+1\right)^3\)

\(\Rightarrow\left\{{}\begin{matrix}y=x+1\\y^3=x^3+3x^2+3x+1\end{matrix}\right.\)

Do \(y^3=x^3+2x^2+3x+2\)

\(\Rightarrow x^3+2x^2+3x+2=x^3+3x^2+3x+1\)

\(\Rightarrow1=x^2\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=1+1=2\\y=-1+1=0\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=1,y=2\\x=-1,y=0\end{matrix}\right.\)

Giả sử điều cần c/m là đúng . Khi đó , ta có :

\(\dfrac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\dfrac{1}{a^2+b^2+c^2}\)

\(\Leftrightarrow\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)=\left(ax+by+cz\right)^2\)

\(\Leftrightarrow x^2a^2+y^2a^2+z^2a^2+x^2b^2+y^2b^2+z^2b^2+x^2c^2+y^2c^2+z^2c^2\)

\(=x^2a^2+b^2y^2+c^2z^2+2axby+2bycz+2axcz\)

\(\Leftrightarrow y^2a^2+z^2a^2+x^2b^2+z^2b^2+x^2c^2+y^2c^2=2axby+2bycz+2axcz\)

\(\Leftrightarrow y^2a^2+z^2a^2+x^2b^2+z^2b^2+x^2c^2+y^2c^2-2axby-2bycz-2axcz=0\) \(\Leftrightarrow\left(y^2a^2-2axby+b^2x^2\right)+\left(b^2z^2-2bycz+c^2y^2\right)+\left(x^2c^2-2axcz+a^2z^2\right)=0\)

\(\Leftrightarrow\left(ay-bx\right)^2+\left(bz-cy\right)^2+\left(cx-az\right)^2=0\left(1\right)\)

Do \(\left\{{}\begin{matrix}\left(ay-bx\right)^2\ge0\\\left(bz-cy\right)^2\ge0\\\left(cx-az\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow\left(ay-bx\right)^2+\left(bz-cy\right)^2+\left(cx-az\right)^2\ge0\left(2\right)\)

Từ ( 1 ) ; ( 2 ) \(\Rightarrow\left\{{}\begin{matrix}ay-bx=0\\bz-cy=0\\cx-az=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}ay=bx\\bz=cy\\cx=az\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{x}=\dfrac{b}{y}\\\dfrac{b}{y}=\dfrac{c}{z}\\\dfrac{c}{z}=\dfrac{a}{x}\end{matrix}\right.\) \(\Rightarrow\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)

Điều này đúng với GT đề bài cho

\(\Rightarrow\) Điều cần c/m là đúng

\(\Rightarrow\dfrac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\dfrac{1}{a^2+b^2+c^2}\left(đpcm\right)\)

\(A=\dfrac{3x^2+5xy-2y^2}{3x^2-7xy+2y^2}\)

\(=\dfrac{3x^2-xy+6xy-2y^2}{3x^2-xy-6xy+2y^2}\)

\(=\dfrac{x\left(3x-y\right)+2y\left(3x-y\right)}{x\left(3x-y\right)-2y\left(3x-y\right)}\)

\(=\dfrac{\left(x+2y\right)\left(3x-y\right)}{\left(x-2y\right)\left(3x-y\right)}\)

\(=\dfrac{x+2y}{x-2y}\)

Sai đề

1 ) Nước chảy đá mòn có nghĩa là nước đã tác dụng vào đá 1 lực đẩy làm cho đá bị mòn ( biến dạng ) . Nước chỉ tác dụng lên hòn đá khi hòn đá đứng yên ở vị trí cố định và để làm đá mòn thì nước phải tác dụng lên đá rất nhiều thời gian

2 ) a Khi xe chuyển động thẳng đều , lực kéo là 1000 N

-> Độ lớn của lực ma sát lên bánh xe đang lăn đều trên đường là

2000 - 1000 = 1000 ( N )

Đ/s : 1000 N

b Hợp lực làm ô tô chạy nhanh dần khi khởi hành là :

\(F_{Hl}+F_{MSL}=F_K\)

-> \(F_{HL}=F_K-F_{MSL}=2000-1000=1000\left(N\right)\)

Đ/s : \(1000N\)

Xe đang chuyển động thẳng đều , đột ngột rẽ trái , do quán tính , người trên xe không thể đổi hướng chuyển động ngay mà tiếp tục theo chuyển động cũ nên sẽ nghiêng người sang phải

\(C=x^2-2xy-4x+2y^2-8y+20\)

\(=\left(x^2-2xy+y^2\right)-4\left(x-y\right)+4+y^2-12y+36-20\)

\(=\left(x-y\right)^2-4\left(x-y\right)+4+\left(y-6\right)^2-20\)

\(=\left(x-y-2\right)^2+\left(y-6\right)^2-20\ge-20\forall x;y\)

Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-y-2=0\\y-6=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=2\\y=6\end{matrix}\right.\)

\(\Leftrightarrow x=8;y=6\)

Vậy Min C là : \(-20\Leftrightarrow x=8;y=6\)

1 ) Ta có : \(2n^2+3n+3\)

\(=2n^2-n+4n-2+5\)

\(=n\left(2n-1\right)+2\left(2n-1\right)+5\)

\(=\left(n+2\right)\left(2n-1\right)+5\)

Để \(2n^2+3n+3⋮2n-1\)

\(\Leftrightarrow\left(n+2\right)\left(2n-1\right)+5⋮2n-1\)

\(\Leftrightarrow5⋮2n-1\)

Do \(n\in Z\Rightarrow2n-1\in Z\)

\(\Rightarrow2n-1\in\left\{1;-1;5;-5\right\}\)

\(\Rightarrow2n\in\left\{2;0;6;-4\right\}\)

\(\Rightarrow n\in\left\{1;0;3;-2\right\}\)

Vậy ...

Bài 2 :

a ) \(3x^2-3y^2+4x-4y=3\left(x^2-y^2\right)+4\left(x-y\right)=3\left(x-y\right)\left(x+y\right)+4\left(x-y\right)\)

\(=\left(x-y\right)\left[3\left(x+y\right)+4\right]=\left(x-y\right)\left(3x+3y+4\right)\)

b ) \(12x^2-3xy+8x-2y\)

\(=3x\left(4x-y\right)+2\left(4x-y\right)\)

\(=\left(3x+2\right)\left(4x-y\right)\)

c ) \(x^3+x^2y-x^2z-xyz\)

\(=x^2\left(x+y\right)-xz\left(x+y\right)\)

\(=\left(x^2-xz\right)\left(x+y\right)\)

\(=x\left(x-z\right)\left(x+y\right)\)

d ) \(xy+y-2x-2\)

\(=y\left(x+1\right)-2\left(x+1\right)\)

\(=\left(y-2\right)\left(x+1\right)\)

e ) \(x^3-3x^2+3x-9\)

\(=x^2\left(x-3\right)+3\left(x-3\right)\)

\(=\left(x^2+3\right)\left(x-3\right)\)

a ) Gọi thương của phép chia \(2x^2+ax+1\) cho \(x-3\) là \(A\left(x\right)\)

\(\Rightarrow2x^2+ax+1=\left(x-3\right)A\left(x\right)+4\)

Chọn \(x=3\) thay vào b/t trên , ta được :

\(2.3^2+3a+1=4\)

\(\Rightarrow18+3a+1=4\)

\(\Rightarrow19+3a=4\)

\(\Rightarrow3a=-15\)

\(\Rightarrow a=-5\)

Vậy \(a=-5\) thì \(2x^2+ax+1\) chia \(x-3\) dư 4

a ) \(x^2-11x-26=0\)

\(\Leftrightarrow x^2-13x+2x-26=0\)

\(\Leftrightarrow x\left(x-13\right)+2\left(x-13\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-13\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=13\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=-2\\x=13\end{matrix}\right.\)

b ) \(2x^2+7x-4=0\)

\(\Leftrightarrow2\left(x^2+\dfrac{7}{2}x-2\right)=0\)

\(\Leftrightarrow x^2+\dfrac{7}{2}x-2=0\)

\(\Leftrightarrow x^2+\dfrac{7}{2}x+\dfrac{49}{16}-\dfrac{81}{16}=0\)

\(\Leftrightarrow\left(x+\dfrac{7}{4}\right)^2=\dfrac{81}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{7}{4}=\dfrac{9}{4}\\x+\dfrac{7}{4}=-\dfrac{9}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-4\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-4\end{matrix}\right.\)

c ) \(\left(x-2\right)\left(x-3\right)+\left(x-2\right)-1=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)+x-3=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-2+1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

\(x^4-5x^2+4\)

\(=x^4-4x^2+4-x^2\)

\(=\left(x^2-2\right)^2-x^2\)

\(=\left(x^2-2-x\right)\left(x^2-2+x\right)\)