\(\left(x+1\right)\left(x-1\right)^2-\left(x+1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)\left[\left(x-1\right)^2-\left(x-2\right)^2\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1-x+2\right)\left(x-1+x-2\right)=0\)
\(\Leftrightarrow\left(x+1\right).1.\left(2x-3\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x-3\right)=0\)
\(TH1:x+1=0\)
\(\Leftrightarrow x=-1\)
\(TH2:2x-3=0\)
\(\Leftrightarrow2x=3\)
\(\Leftrightarrow x=\frac{3}{2}\)
Vậy phương trình có tập nghiệm là:\(S=\left\{-1;\frac{3}{2}\right\}\)
\(\left(x+1\right)\left(x-1\right)^2-\left(x+1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)\left[\left(x-1\right)^2-\left(x-2\right)^2\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left[\left(x-1-x+2\right)\left(x-1+x-2\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left[1.\left(2x-3\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\2x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\2x=3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{3}{2}\end{cases}}\)
Vậy................
