a) Ta có: \(2\left(2x-1\right)^2-\left(x-3\right)^2=\left(x+1\right)\left(7x+2\right)\)
\(\Rightarrow2\left(4x^2-4x+1\right)-\left(x^2-6x+9\right)=\left(x+1\right)\left(7x+2\right)\)
\(\Rightarrow8x^2-8x+2-x^2+6x-9=\left(x+1\right)\left(7x+2\right)\)
\(\Rightarrow7x^2-2x-7=7x^2+2x+7x+2\)
\(\Rightarrow7x^2-2x-7-7x^2-2x-7x-2=0\)
\(\Rightarrow-11x-9=0\)
\(\Rightarrow-11x=9\)
hay \(x=\frac{9}{-11}=\frac{-9}{11}\)
Vậy: \(x=\frac{-9}{11}\)
b) Ta có: \(\left(x+1\right)^3=4x+4\)
\(\Rightarrow\left(x+1\right)^3=4\left(x+1\right)\)
\(\Rightarrow\left(x+1\right)^3-4\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left[\left(x+1\right)^2-4\right]=0\)
\(\Rightarrow\left(x+1\right)\left[\left(x+1\right)^2-2^2\right]=0\)
\(\Rightarrow\left(x+1\right)\left[\left(x+1-2\right)\left(x+1+2\right)\right]=0\)
\(\Rightarrow\left(x+1\right)\left(x-1\right)\left(x+3\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x+1=0\\x-1=0\\x+3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\x=1\\x=-3\end{matrix}\right.\)
Vậy: \(x\in\left\{-1;1;-3\right\}\)
a, \(2\left(2x-1\right)^2-\left(x-3\right)^2=\left(x+1\right)\left(7x+2\right)\)
\(2\left(4x^2-4x+1\right)-\left(x^2-6x+9\right)=7x^2+7x+2x+2\)
\(8x^2-8x+2-x^2+6x-9=7x^2+9x+2\)
\(7x^2-2x-7=7x^2+9x+2\)
\(-11x=9\) => \(x=\frac{-9}{11}\)
b,\(\left(x+1\right)^3=4x+4\)
\(\left(x+1\right)^3=4\left(x+1\right)\)
\(\left(x+1\right)^2=4\)
\(x+1=2\)
\(x=1\)
