\(4x^2-4x\left(x-3\right)+\left(3-x\right)^2=0\)

\(\Leftrightarrow\left(2x\right)^2-2.2x\left(x-3\right)+\left(x-3\right)^2=0\)

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

Từ:

 \(\hept{\begin{cases}a^2\left(b+c\right)=2012\\b^2\left(a+c\right)=2012\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}a^2\left(b+c\right)-b^2\left(a+c\right)=0\\b^2\left(a+c\right)+a^2\left(b+c\right)=4024\end{cases}}\)

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

\(\Leftrightarrow ab\left(a-b\right)+c\left(a+b\right)\left(a-b\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(ab+bc+ca\right)=0\)

\(\Leftrightarrow ab+bc+ca=0\left(a\ne b\right)\)

\(b^2\left(a+c\right)+a^2\left(b+c\right)=4024\)

\(\Leftrightarrow ab\left(a+b\right)+c\left(a+b\right)^2-2abc=4024\)

\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ca\right)-2abc=4024\)

\(\Leftrightarrow2abc=-4024\)

\(\Leftrightarrow abc=-2012\)

\(ab\left(a-b\right)+c\left(a+b\right)\left(a-b\right)=0\)

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

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

\(\Leftrightarrow c^2\left(a+b\right)=-abc=2012\)

(x - 2)³ - x³ + 6x² = 5

<=> x³ - 6x² + 12x - 8 - x³ + 6x² = 5

<=> 12x - 8 = 5

<=> 12x = 13

<=> \(x=\frac{13}{12}\)

NHA

MÌNH MÁY TÍNH  KO RÕ NÊN KO THẤY CÂU HỎI CỦA YOU

| x + 3 | = 2x - 1

\(\Rightarrow\orbr{\begin{cases}x+3=2x-1\\x+3=-(2x-1)=1-2x\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x-x=3+1\\x+2x=1-3\end{cases}\Rightarrow}\orbr{\begin{cases}x=4\\3x=-2\end{cases}\Rightarrow}\orbr{\begin{cases}x=4\\x=\frac{-2}{3}\end{cases}}\)

Cách đó chỉ giải với 2 vế là GTTĐ thôi nhá @Sơn 

CTTQ : \(\left|f\left(x\right)\right|=g\left(x\right)\)ĐK : \(g\left(x\right)\ge0\)

TH1 : \(f\left(x\right)=g\left(x\right)\)

TH2 : \(f\left(x\right)=-g\left(x\right)\)

-> rồi xét x có thỏa mãn với đk ko rồi kết luận 

\(\left|x+3\right|=2x-1\)ĐK : \(x\ge\frac{1}{2}\)

TH1 : \(x+3=2x-1\Leftrightarrow x=4\)(tmđk )

TH2 : \(x+3=1-2x\Leftrightarrow3x=-2\Leftrightarrow x=-\frac{2}{3}\)(ktmđk) 

Vậy x = 4 

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

$\iff \frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}$

$\iff x^2+2x-x+2-2=0$

$\iff x^2+x=0$

$\iff x\left(x+1\right)=0\iff [\begin{array}{c}x=-1\\ x=0\end{array}$

Phân số mà các bạn

x⁴ - ( 4x - 4 )² = ( x² - 4x + 4 )( x² + 4x - 4 ) = ( x - 2 )²( x² + 4x - 4 )

\(3=a^2+ab+b^2=\left(a+b\right)^2-ab\Leftrightarrow ab=\left(a+b\right)^2-3\ge-3\)

Dấu \(=\)khi \(\orbr{\begin{cases}x=\sqrt{3},y=-\sqrt{3}\\x=-\sqrt{3},y=\sqrt{3}\end{cases}}\).

\(3=a^2+ab+b^2=\left(a-b\right)^2+3ab\Leftrightarrow ab=\frac{3-\left(a-b\right)^2}{3}\le1\)

Dấu \(=\)khi \(\orbr{\begin{cases}x=y=1\\x=y=-1\end{cases}}\).