Ôn tập toán 8

a)x³-4x²+8x-8

=x³-2x²+4x-2x²+4x-8

=x(x²-2x+4)-2(x²-2x+4)

=(x-2)(x²-2x+4)

b)x³-5x²-5x+1

=x³-6x²+x+x²-6x+1

=x(x²-6x+1)+(x²-6x+1)

=(x+1)(x²-6x+1)

c)x²y²-x²+8x-16

=x²y²+x²y-4xy-x²y-x²+4x+4xy+4x-16

=xy(xy+x-4)-x(xy+x-4)+4(xy+x-4)

=(xy-x+4)(xy+x-4)

d)x²-6xy+9y²+4x-12y

=9y²-3xy-12y-3xy+x²+4x

=3y(3y-x-4)-x(3y-x-4)

=(3y-x-4)(3y-x)

e)x²-xy+x³-3x²y+3xy²-y³

=x(x-y)+(x-y)³

=(x-y)[x+(x-y)²]

=(x-y)(x+x²-2xy+y²)

f)x⁴-x³+x²-1

=x⁴+x²+x-x³-x-1

=x(x³+x+1)-(x³+x+1)

=(x-1)(x³+x+1)

g)x⁴-x²+10x-25

=x⁴+x³-5x²-x³-x²+5x+5x²+5x-25

=x²(x²+x-5)-x(x²+x-5)+5(x²+x-5)

=(x²-x+5)(x²+x-5)

Đặt \(x=\frac{a}{b}+\frac{b}{a}\Rightarrow\frac{a^2}{b^2}+\frac{b^2}{a^2}=x^2-2\)

Xét mẫu thức : \(\frac{a^2}{b^2}+\frac{b^2}{a^2}-\left(\frac{a}{b}+\frac{b}{a}\right)=x^2-x-2=\left(x+1\right)\left(x-2\right)\)

Thay \(x=\frac{a}{b}+\frac{b}{a}\) được mẫu thức : \(\left(\frac{a}{b}+\frac{b}{a}+1\right)\left(\frac{a}{b}+\frac{b}{a}-2\right)=\left(\frac{a}{b}+\frac{b}{a}+1\right).\frac{\left(a-b\right)^2}{ab}\)

Ta có : \(P=\frac{\left(\frac{a}{b}+\frac{b}{a}+1\right)\left(\frac{1}{a}-\frac{1}{b}\right)^2}{\frac{a^2}{b^2}+\frac{b^2}{a^2}-\left(\frac{a}{b}+\frac{b}{a}\right)}=\frac{\left(\frac{a}{b}+\frac{b}{a}+1\right).\frac{\left(a-b\right)^2}{a^2b^2}}{\left(\frac{a}{b}+\frac{b}{a}+1\right).\frac{\left(a-b\right)^2}{ab}}\)

\(=\frac{\left(a-b\right)^2}{a^2b^2}.\frac{ab}{\left(a-b\right)^2}=\frac{1}{ab}\) (đpcm)

b) Áp dụng bđt Cauchy : 

\(1=4a+b+\sqrt{ab}\ge2\sqrt{4a.b}+\sqrt{ab}\)

\(\Rightarrow5\sqrt{ab}\le1\Rightarrow ab\le\frac{1}{25}\)

\(\Rightarrow P=\frac{1}{ab}\ge25\) . Dấu "=" xảy ra khi \(\begin{cases}4a+b+\sqrt{ab}=1\\4a=b\end{cases}\)

\(\Leftrightarrow\begin{cases}a=\frac{1}{10}\\b=\frac{2}{5}\end{cases}\) 

Vậy P đạt giá trị nhỏ nhất bằng 25 tại \(\left(a;b\right)=\left(\frac{1}{10};\frac{2}{5}\right)\)

Có: \(P=A:B=\left(\frac{1}{x-1}-\frac{2x}{x^3+x-x^2-1}\right):\left(1-\frac{2x}{x^2+1}\right)\left(ĐK:x\ne1\right)\)

\(=\left[\frac{1}{x-1}-\frac{2x}{x\left(x^2+1\right)-\left(x^2+1\right)}\right]:\left(\frac{x^2+1-2x}{x^2+1}\right)\)

\(=\left[\frac{1}{x-1}-\frac{2x}{\left(x^2+1\right)\left(x-1\right)}\right]:\frac{\left(x-1\right)^2}{x^2+1}\)

\(=\frac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}\cdot\frac{x^2+1}{\left(x-1\right)^2}\)

\(=\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x^2+1\right)}\cdot\frac{x^2+1}{\left(x-1\right)^2}=\frac{1}{x-1}\)

b) Để \(P>-\frac{1}{2}\)

\(\Leftrightarrow\)\(\frac{1}{x-1}>-\frac{1}{2}\)

\(\Leftrightarrow\)\(\frac{1}{x-1}+\frac{1}{2}>0\)

\(\Leftrightarrow\)\(\frac{2+x-1}{2\left(x-1\right)}>0\)

\(\Leftrightarrow\)\(\frac{x+1}{2\left(x-1\right)}>0\)

\(\Leftrightarrow\begin{cases}x+1>0\\2\left(x-1\right)>0\end{cases}\) hoặc \(\begin{cases}x+1< 0\\2\left(x-1\right)< 0\end{cases}\)

\(\Leftrightarrow\begin{cases}x>-1\\x>1\end{cases}\) hoặc \(\begin{cases}x< -1\\x< 1\end{cases}\)

\(\Leftrightarrow x>1\) hoặc \(x< -1\)

Điều kiện : \(a\ne1\)

\(A=\left(1+\frac{a}{a^2+1}\right):\left(\frac{1}{a-1}+\frac{2a}{a^2+1-a^3-a}\right)-1\)

\(=\frac{a^2+a+1}{a^2+1}:\left(\frac{-a^2-1}{\left(1+a^2\right)\left(1-a\right)}+\frac{2a}{\left(1+a^2\right)\left(1-a\right)}\right)-1\)

\(=\frac{a^2+a+1}{a^2+1}.\frac{\left(a-1\right)\left(1+a^2\right)}{\left(a-1\right)^2}-1=\frac{a^2+a+1}{a-1}-1=\frac{a^2+2}{a-1}\)

b) A < 2 \(\Rightarrow\frac{a^2+2}{a-1}< 2\Leftrightarrow\frac{\left(a^2-2a+1\right)+2\left(a-1\right)+3}{a-1}< 2\)

\(\Leftrightarrow a-1+2+\frac{3}{a-1}< 2\Leftrightarrow a-1+\frac{3}{a-1}< 0\)

Đặt t = a-1 , xét : 

Nếu t > 0 thì \(t+\frac{3}{t}< 0\Leftrightarrow t^2+3< 0\) không thỏa mãn vì \(t^2+3>3>0\)

Nếu t < 0 thì \(t+\frac{3}{t}< 0\Leftrightarrow t^2+3>0\) thỏa mãn

Vậy a - 1 < 0 => a < 1 thỏa mãn đề bài

vì EA vuông góc với OM (gt)

    BF vuông góc với OM (gt)

nên AE // BF→ góc EAO = góc OBF

Xét tam giác AEO và tam giác OBF có 

    góc AOE =góc BOF (đối đỉnh )

    góc EAO = góc OBF (cmt)

    AO = OB (gt)

→ΔAEO=ΔBFO(g.c.g)

→AE=BF(đpcm)

nếu đúng tích hộ mình nhá

\(2x+3=0\Leftrightarrow2x=-3\Leftrightarrow x=-\frac{3}{2}\)

=> 2x = - 3

\(\Rightarrow x=-\frac{3}{2}\)

2x+3=0 

2x = 0 - 3

2x = -3

x = -3 : 2

x = -1,5

Đặt \(k^2=n^2+31n+1984\) (k thuộc N)

Ta có \(n^2+30n+225< n^2+31n+1984< n^2+90n+2025\)

\(\Rightarrow\left(n+15\right)^2< k^2< \left(n+45\right)^2\)

Xét k² trong khoảng trên được n = 565 và n = 1728 thỏa mãn đề bài.

17) \(8x^3+27=\left(2x\right)^3+3^3=\left(2x+3\right)\left(4x^2-6x+9\right)\)

18) \(a^6-1=\left(a^3\right)^2-1=\left(a^3-1\right)\left(a^3+1\right)=\left(a-1\right)\left(a^2+a+1\right)\left(a+1\right)\left(a^2-a+1\right)\)

20) \(9x^4-81x^2=9\left(x^4-9x^2\right)=9x^2\left(x^2-9\right)=9x^2\left(x-3\right)\left(x+3\right)\)

17)

\(8x^3+27=\left(8x\right)^3+3^3=\left(8x+3\right)\left(64x^2-24x+9\right)\)

18)

\(a^6-1=\left(a^3\right)^2-1\)

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

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

19)

đề sao sao ý

20)

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

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

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

Bài 2:

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

\(=\left(x-1-x-1\right)^3=\left(-2\right)^3=-8\)

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

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

\(=\left[\left(1-x\right)+\left(1+x\right)\right]^3=2^3=8\)

cho mình sửa 1 chỗ ,  là (1-x)³ + 3( x² - 1 ) ( x +1 ) + 3(x² - 1) ( x+1) +(x+1)³ 

xl mọi người đừng làm câu này để mik ghi lại đề