\(\left(x^2-4\right)\left(2x+3\right)=\left(x^2-4\right)\left(x-1\right)\\ \Rightarrow2x^3+3x^2-8x-12=x^3-x^2-4x+4\\ \Rightarrow x^3+4x^2-4x-16=0\\ \Rightarrow\left(x^3-4x\right)+\left(4x^2-16\right)=0\\ \Leftrightarrow x\left(x^2-4\right)+4\left(x^2-4\right)=0\\ \Leftrightarrow\left(x^2-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\pm2\\x=-4\end{matrix}\right.\)
\(\left(3x-7\right)^2-4\left(x+1\right)^2=0\\ \Leftrightarrow\left(3x-7\right)^2-\left(2x+2\right)^2=0\\ \Leftrightarrow\left(5x-5\right)\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=9\end{matrix}\right.\)
a)(x2-4)(2x+3)=(x2-4)(x-1)
=>2x+3 = x-1(cả hai đều chia cho x2-4)
=>2x+3-x+1=0
=>x+4 =0
=> x = -4
Vậy S={-4}
b)(3x-7)2-4(x+1)2=0
=> (3x-7)2-[2(x+1)]2=0
=>[(3x-7)-2(x+1)][(3x-7)+2(x+1)]=0
=>(3x-7-2x-2)(3x-7+2x+2)=0
=>(x-9)(5x-5)=0
=>5(x-9)(x-1)=0
=> x-9=0 và x-1=0
<=> x = 9 và x=1
Vậy S={1;9}
\(\text{a) }\left(x^2-4\right)\left(2x+3\right)=\left(x^2-4\right)\left(x-1\right)\\ \Leftrightarrow\left(x^2-4\right)\left(2x+3\right)-\left(x^2-4\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-4\right)\left(2x+3-x+1\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-2\right)\left(x+4\right)\\ \Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-2=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=2\\x=-4\end{matrix}\right.\\ \text{Vậy }S=\left\{-2;2;-4\right\}\)
\(\text{b) }\left(3x-7\right)^2-4\left(x+1\right)^2=0\\ \Leftrightarrow\left(3x-7\right)^2-\left(2x+2\right)^2=0\\ \Leftrightarrow\left(3x-7-2x-2\right)\left(3x-7+2x+2\right)=0\\ \Leftrightarrow\left(x-9\right)\left(5x-5\right)=0\\ \Leftrightarrow5\left(x-9\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-9\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-9=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\\ \\ \text{Vậy }S=\left\{9;1\right\}\)