a/ \(x^3-3x^2+3x-2=0\)
\(\Leftrightarrow x^3-2x^2-x^2+2x+x-2=0\)
\(\Leftrightarrow x^2\left(x-2\right)-x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x^2-x+1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=0\left(ktm\right)\end{cases}}\)
Vậy x = 2 là nghiệm của phương trình.
b/ \(\left(x+y\right)^2=\left(x-1\right)\left(y+1\right)\)
\(\Leftrightarrow2\left(x+y\right)^2=2\left(x-1\right)\left(y+1\right)\)
\(\Leftrightarrow2x^2+4xy+2y^2=2xy+2x-2y-2\)
\(\Leftrightarrow2x^2+2y^2+2xy-2x+2y+2=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+2y+1\right)+\left(x^2+2xy+y^2\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+1\right)^2+\left(x+y\right)^2=0\)
Mà \(\left(x-1\right)^2\ge0\)
\(\left(y+1\right)^2\ge0\)
\(\left(x+y\right)^2\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x=1\\y=-1\end{cases}}\)
Vậy \(x=1;y=-1\Leftrightarrow\left(x+y\right)^2=\left(x-1\right)\left(y+1\right)\)
