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

\(\Leftrightarrow\left(2x-5+x+2\right)\left(2x-5-x-2\right)=0\)

\(\Leftrightarrow\left(3x-3\right)\left(x-7\right)=0\)

b) Cách làm giống câu a

\(\left(3x^2+10x-8\right)^2=\left(5x^2-2x+10\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}3x^2+10x-8=5x^2-2x+10\\3x^2+10x-8=-5x^2+2x-10\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x^2-12x+18=0\\8x^2+8x+2=0\end{cases}}\)

\(TH1:2x^2-12x+18=0\)

\(\Leftrightarrow x^2-6x+9=0\)

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

\(TH2:8x^2+8x+2=0\)

\(\Leftrightarrow4x^2+4x+1=0\)

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

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

TH1: 5x² - 2x + 10 = 3x² + 10x - 8

      => 2x² - 12x + 18 = 0

      => x² - 6x + 9 = 0

      => (x - 3)² = 0

      => x = 3

TH2: 5x² - 2x + 10 = - 3x² - 10x + 8

      => 8x² + 8x + 2 = 0

      => 4x² + 4x + 1 = 0

      => (2x + 1)² = 0 

      => x = -1/2

Vậy x = 3 , x = -1/2

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

hay \(x\in\left\{-\dfrac{2}{3};\dfrac{-1+\sqrt{17}}{4};\dfrac{-1-\sqrt{17}}{4}\right\}\)

c: \(\Leftrightarrow\left[{}\begin{matrix}3x^2+10x+21=x^2-20x-9\\3x^2+10x+21=-x^2+20x+9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x^2+30x+30=0\\4x^2-10x+12=0\end{matrix}\right.\Leftrightarrow x\in\left\{\dfrac{-15+\sqrt{165}}{2};\dfrac{-15-\sqrt{165}}{2}\right\}\)

`(3x^2+10x-8)=(5x^2-2x+10)^2`

`<=>(3x^2+10x-8+5x^2-2x+10)(3x^2+10x-8-5x^2+2x-10)=0`

`<=> (8x^2+8x+2)(-2x^2+12x-18)=0`

\(\Leftrightarrow\left[{}\begin{matrix}8x^2+8x+2=0\\-2x^2+12x-18=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)

Vậy `S={-1/2 ; 3}`.

=>( 2x² -12x +18 ) ( 8x² +8x +2) =0

=> x² - 6x + 9 =0 => x =3

hoặc 4x² +4x +1 =0 =>x =-1/2 

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