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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-1=0\\x^2-x-2=0\end{matrix}\right.\)

a. \(\left(2x-3\right)\left(x+1\right)+\left(2x-3\right)\left(3x-7\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(x+1+3x-7\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(4x-6\right)=0\)

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

b. \(\left(x-4\right)\left(3x-2\right)+x^2-16=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\4x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{2}\end{matrix}\right.\)

(2x-3)(x+1)+(2x+3)(3x-7)=0

<=> (2x-3)(x+1)-(2x-3)(3x-7)=0

<=> (2x-3)(x+1-3x+7)=0

<=> (2x-3)(-2x+8)=0

<=> 2x-3=0 => x=3/2

Hoặc -2x+8=0 => x= 4

Vậy x thuộc{3/2;4}

\(A=3\left(2x-3\right)\left(3x+2\right)-\left(2x+4\right)\left(4x-3\right)+9x\left(4-x\right)\)

\(=\left(6x-9\right)\left(3x+2\right)-8x^2+6x-16x+12+36x-9x^2\)

\(=18x^2+12x-27x-18-17x^2+26x+12\)

\(=x^2+11x-6\)

Để A = 0

\(\Leftrightarrow x^2+11x-6=0\)

\(\Leftrightarrow\left(x^2+11x+\frac{121}{4}\right)-\frac{145}{4}=0\)

\(\Leftrightarrow\left(x+\frac{11}{2}\right)^2-\left(\frac{\sqrt{145}}{2}\right)^2=0\)

\(\Leftrightarrow\left(x+\frac{11}{2}-\frac{\sqrt{145}}{2}\right)\left(x+\frac{11}{2}+\frac{\sqrt{145}}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{145}-11}{2}\\x=\frac{-\sqrt{145}-11}{2}\end{matrix}\right.\)

Vậy..................

Bài 2: 

a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

Đăng từng câu đio

,(8x-3).(3x+2)-(4x+7).(x+4)=(2x+1).(5x-1)-33 đúng không bạn

     \(x^4-2x^3-2x^2+3x+2=0\)

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

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

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

\(\Leftrightarrow\left(x-2\right)\left(x^3-2x-1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3-x-x-1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left[\left(x^3-x\right)-\left(x+1\right)\right]=0\)

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

\(\Leftrightarrow\left(x-2\right)\left[x\left(x-1\right)\left(x+1\right)-\left(x+1\right)\right]=0\)

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

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

Đến đây ez r