\(5,\left(x+1\right)^2=4\left(x^2-2x+1\right)\)
\(\Leftrightarrow\left(x+1\right)^2-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)^2-\left[2\left(x-1\right)\right]^2=0\)
\(\Leftrightarrow\left(x+1\right)^2-\left(2x-2\right)^2=0\)
\(\Leftrightarrow\left(x-1-2x+2\right)\left(x-1+2x-2\right)=0\)
\(\Leftrightarrow\left(-x+1\right)\left(3x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x+1=0\\3x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=1\end{matrix}\right.\)
Vậy pt có nghiệm duy nhất x = 1
\(6,2x^3+5x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow x\left(2x^2+6x-x-3\right)=0\)
\(\Leftrightarrow x\left[2x\left(x+3\right)-\left(x+3\right)\right]=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
