a) \(\Leftrightarrow x^2-x-x^2+2x=5\)
    \(\Leftrightarrow x=5\)
b) \(\Leftrightarrow4x\left(x^2-9\right)=0\)
    \(\Leftrightarrow4x\left(x-3\right)\left(x+3\right)=0 \)
    \(\Leftrightarrow\)\(\left[{}\begin{matrix}4x=0\\x-3=0\\x+3=0\end{matrix}\right.\)
    \(\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
Vậy x = 0 , x = 3 hoặc x = -3

\(a,\Leftrightarrow x^2-x-x^2+2x=5\\ \Leftrightarrow x=5\\ b,\Leftrightarrow4x\left(x^2-9\right)=0\\ \Leftrightarrow4x\left(x-3\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow2x\left(x-1\right)-\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(2x-x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\\ d,\Leftrightarrow\left(x^2-9x+14\right)\left(x^2-9x+20\right)-72=0\\ \Leftrightarrow\left(x^2-9x+17\right)^2-3^2-72=0\\ \Leftrightarrow\left(x^2-9x+17\right)^2-81=0\\ \Leftrightarrow\left(x^2-9x+17-9\right)\left(x^2-9x+17+9\right)=0\\ \Leftrightarrow\left(x-8\right)\left(x-1\right)\left(x^2-9x+26\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=1\\\left(x-\dfrac{9}{2}\right)^2+\dfrac{23}{4}=0\left(vô.n_0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=8\end{matrix}\right.\)

bai toan nay kho

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

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

(x-2)*(1+2X+1)=0

vay:x=2

    x=-1

suy ra (x-2)(2x+1)=2-x=-(x-2)

<=>(x-2)(2x+1)+(x-2)=0

<=>(x-2)(2x+1)=0

<=>x-2=0  hoặc 2x+1=0

<=> x=2 hoặc x=-1/2

Vậy tập nghiệm của phương trình là .........

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