Những hằng đẳng thức đáng nhớ - Hỏi đáp

Ta có: \(x^2-4x+9\)

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

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

Ta có: \(\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-2\right)^2+5\ge5\forall x\)

Dấu '=' xảy ra khi

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

Vậy: GTNN của đa thức \(x^2-4x+9\) là 5 khi x=2

Lời giải:

Ta có: \(f(x)=x^2-4x+9=(x^2-4x+4)+5=(x-2)^2+5\)

Vì \((x-2)^2\geq 0, \forall x\in\mathbb{R}\)

\(\Rightarrow f(x)=(x-2)^2+5\geq 5\)

Vậy GTNN của đa thức là $5$. Giá trị này đạt tại $(x-2)^2=0\Leftrightarrow x=2$

Bài 3:

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{ay-bx}{c}=\frac{cx-az}{b}=\frac{bz-cy}{a}\)

\(\Rightarrow \frac{ayz-bxz}{cz}=\frac{cxy-azy}{by}=\frac{bzx-cyx}{ax}=\frac{ayz-bxz+cxy-azy+bzx-cyx}{cz+by+ax}=0\)

\(\Rightarrow \left\{\begin{matrix} ayz-bxz=0\\ cxy-azy=0\\ bzx-cyx=0\end{matrix}\right.\Rightarrow ayz=bxz=cxy\)

\(\Rightarrow \frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)

Đặt \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=t\Rightarrow a=xt; b=yt; c=zt\)

Khi đó:

\((ax+by+cz)^2=(x^2t+y^2t+z^2t)^2=t^2(x^2+y^2+z^2)^2\)

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

Từ đây ta có đpcm.

Bài 1:

\(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow (a+b)^3-3ab(a+b)+c^3=3abc\)

\(\Leftrightarrow (a+b)^3+c^3-3ab(a+b+c)=0\)

\(\Leftrightarrow (a+b+c)[(a+b)^2-(a+b)c+c^2]-3ab(a+b+c)=0\)

\(\Leftrightarrow (a+b+c)[(a+b)^2-(a+b)c+c^2-3ab]=0\)

\(\Leftrightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0\)

Vì $a,b,c$ dương nên $a+b+c\neq 0$

Do đó $a^2+b^2+c^2-ab-bc-ac=0$

$\Leftrightarrow \frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}=0$

$\Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2=0$

Do $(a-b)^2; (b-c)^2; (c-a)^2\geq 0$ với mọi $a,b,c>0$

Suy ra để tổng của chúng bằng $0$ thì $(a-b)^2=(b-c)^2=(c-a)^2=0$

$\Rightarrow a=b=c$ (đpcm)

Sửa lại đề: \(a+b+c=0\)

a) Ta có:

\(A=a^4+b^4+c^4=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)\)

\(=[(a+b+c)^2-2(ab+bc+ac)]^2-2(a^2b^2+b^2c^2+c^2a^2)\)

\(=4(ab+bc+ac)^2-2(a^2b^2+b^2c^2+c^2a^2)\)

\(=4(ab+bc+ac)^2-2(a^2b^2+b^2c^2+c^2a^2)-4abc(a+b+c)\)

(do \(a+b+c=0\))

\(A=4(ab+bc+ac)^2-2[a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c)]\)

\(=4(ab+bc+ac)^2-2(ab+bc+ac)^=2(ab+bc+ac)^2\)

Ta có đpcm

b) Ta có:

\(\frac{(a^2+b^2+c^2)^2}{2}=\frac{[(a+b+c)^2-2(ab+bc+ac)]^2}{2}=\frac{[-2(ab+bc+ac)]^2}{2}=2(ab+bc+ac)^2\)

Kết hợp với kết quả phần a ta có đpcm.

\(x^2+4y^2+13-6x+8y=0\)

\(\Rightarrow x^2+4y^2+9+4-6x+8y=0\)

\(\Rightarrow x^2-6x+9+4y^2+8y+4=0\)

\(\Rightarrow\left(x^2-6x+9\right)+\left(4y^2+8y+4\right)=0\)

\(\Rightarrow\left(x^2-2.x.3+3^2\right)+\left[\left(2y\right)^2+2.2y.2+2^2\right]=0\)

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

\(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^2=0\\\left(2y+2\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-3=0\\2y+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0+3\\2y=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)

Vậy \(x=3;y=-1.\)

Chúc bạn học tốt!

Đặt : \(A=x^2+x+1\)
=> \(A=x^2+2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2+1-\left(\frac{1}{2}\right)^2\)

=> \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)

Vì \(\left(x+\frac{1}{2}\right)^2\ge0,\forall x\)
=> \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4},\forall x\)
=> \(A\ge\frac{3}{4},\forall x\)
=> A > 0, \(\forall x\)

Vậy : A > 0

Ta có:

\(x^2+x+1\\ =\left(x^2+2.x.\frac{1}{2}+\frac{1}{4}\right)+\frac{3}{4}\\ =\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)

Vì \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)

=> ĐPCM

#Nguồn: Băng

Sửa lại đề: Tìm GTLN

\(B=-\left(x^2+2xy+y^2\right)+2\left(x+y\right)-y^2-2y-15\)

\(B=-\left[\left(x+y\right)^2-2\left(x+y\right)+1\right]-\left(y^2+2y+1\right)-13\)

\(B=-\left(x+y-1\right)^2-\left(y+1\right)^2-13\)

Do \(-\left(y+1\right)^2\le0\forall y;-\left(x+y-1\right)\le0\forall x,y\)

\(\Rightarrow B\le-13\)

\(\Rightarrow GTLN\) của \(B=-13\) dấu " = " sảy ra:

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

a) 4x² + 4x +1 = 25

⇌ ( 2x+ 1)² - 25 = 0

( 2x + 1 - 5) (2x + 1 + 5) = 0

(2x - 4) (2x + 6) = 0

\(⇌\left[{}\begin{matrix}2x-4=0\\\\2x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\\\x=-3\end{matrix}\right.\)

Hk tốt!

a, 4x² + 4x + 1 = 25

⇔ ( 2x + 1 ) \(^2\) - 25 = 0

⇔( 2x + 1 - 5 ) ( 2x + 1 + 5 ) =0

⇔ ( 2x - 4 ) ( 2x + 6 ) = 0

\(\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\2x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

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

\(2,x^4+13-6x^2+4y+y^2\)

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

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

x³+2x²+2x+1

(x³+1)+(2x²+2x)

(x+1)(x²-x+1)+2x(x+1)

(x+1)(x² -x+1+2x)

(x+1)(x²+x+1)

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

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

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

$x^3+3x^2-x-3$

$\mathrm{=(x3+3x2)-(x+3)}$

$\mathrm{=x2(x+3)-(x+3)}$

$\mathrm{=(x+3)(x2-1)}$

$\mathrm{=(x+3)(x-1)(x+1)}$

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

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

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

a)\(A=x^2+x+1\)

\(A=x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}\)

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

Vậy A_min = 3/4 <=> x = -1/2

b)\(B=2x^2-5x-2\)

\(B=\left(\sqrt{2}x\right)^2-2.\sqrt{2}.\sqrt{2}+\left(\sqrt{2}\right)^2-9\)

\(B=\left(\sqrt{2}x-\sqrt{2}\right)^2-9\ge-9\)

Vậy B_min = -9 <=> x = 1

Nhận thấy \(x=0\) ko phải nghiệm, chia 2 vế cho \(x^2\):

\(2x^2-5x+6-\frac{5}{x}+\frac{2}{x^2}=0\)

\(\Leftrightarrow2\left(x^2+\frac{1}{x^2}\right)-5\left(x+\frac{1}{x}\right)+6=0\)

Đặt \(x+\frac{1}{x}=a\Rightarrow x^2+\frac{1}{x^2}=a^2-2\)

\(2\left(a^2-2\right)-5a+6=0\)

\(\Leftrightarrow2a^2-5a+2=0\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\)

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

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

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

\(=x^3-y^3-x^3-y^3\)

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

\(=-2y^3.\)

Chúc bạn học tốt!

(x^2+xy+y^2)(x-y)-(x+y)(x^2-xy+y^2)=x^3-Y^3-x^3-y^3=-2y^3