Giải các phương trình sau bằng cách đưa về phương trình tích:

$$\begin{gathered} a) (2x+1\left)\right(3x-2)=(5x-8\left)\right(2x+1\left) \\ b\right) 4x^2-1=(2x+1)(3x-5) \\ c) {(x+1)}^2=4(x^2-2x+1\left) \\ d\right) 2x^3+5x^2-3x=0 \end{gathered}$$

Giải

$$\begin{gathered} a) (2x+1\left)\right(3x-2)=(5x-8\left)\right(2x+1) \\ \iff (2x+1\left)\right(3x-2)-(2x+1\left)\right(5x-8)=0 \\ \iff (2x+1)\left[(3x-2)-(5x-8)\right]=0 \\ \iff (2x+1\left)\right(6-2x)=0\iff 2x+1=0 hoac 6-2x=0 \\ *2x+1=0\iff x=-\frac{1}{2} \\ *6-2x=0\iff x=3 \\ \Rightarrow S=\left\{-\frac{1}{2};3\right\} \end{gathered}$$

b) $$\begin{gathered} 4x^2-1=(2x+1)(3x-5)\iff 4x^2-1=6x^2-7x-5 \\ \iff 2x^2-7x-4=0\iff 2x^2-8x+x-4=0 \\ \iff 2x(x-4)+(x-4)=0\iff (x-4)(2x+1)=0 \\ \iff x-4=0 hoac 2x+1=0 \\ \Rightarrow S=\{-\frac{1}{2};4\} \end{gathered}$$

$$\begin{gathered} c) {(x+1)}^2=4(x^2-2x+1)\iff {(x+1)}^2-{\left[2(x-1)\right]}^2=0 \\ \iff \left[(x+1)+2(x-1)\right]\left[(x+1)-2(x-1)\right]=0 \\ \iff (3x-1\left)\right(3-x)=0\iff 3x-1=0 hoac (3-x)=0 \\ \Rightarrow S=\left\{\frac{1}{3};3\right\} \end{gathered}$$

$$\begin{gathered} d) 2x^3+5x^2-3x=0\iff 2x^3+6x^2-x^2-3x=0 \\ \iff 2x^2(x+3)-x(x+3)=0\iff x(x+3\left)\right(2x-1)=0 \\ \iff x=0 hoac x+3=0 hoac 2x-1=0 \\ \Rightarrow S=\left\{-3;0;\frac{1}{2}\right\} \end{gathered}$$