Question

1) 2x²-5x+2=0

2) (x-1)²=2(x²-1)

3) 2(x+2)²-x³-8=0

4) x³-3x²+3x-1=(x-1)(x+1)

Answer 1

1) 2x²-5x+2=0

\(\Leftrightarrow\) 2x2-4x-x+2=0

\(\Leftrightarrow\)2x(x-2)-(x-2)=0

\(\Leftrightarrow\)(x-2)(2x-1)=0

\(\Leftrightarrow\)\(\left[{}\begin{matrix}\text{x-2=0}\\\text{2}x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left[{}\begin{matrix}\text{x=2}\\\text{x}=\frac{1}{2}\end{matrix}\right.\)

vậy tập nghiệm của pt là S={2; \(\frac{\text{1}}{\text{2}}\)}

2) (x-1)²=2(x²-1)

\(\Leftrightarrow\)(x-1)²-2(x-1)(x+1)=0

\(\Leftrightarrow\)(x-1)[(x-1)-2(x+1)]=0

\(\Leftrightarrow\)-(x-1)(x+3)=0

\(\Leftrightarrow\)\(\left[{}\begin{matrix}\text{-(x-1)=0}\\\text{x}+3=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left[{}\begin{matrix}\text{x-1=0}\\\text{x}+3=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left[{}\begin{matrix}\text{x=1}\\\text{x}=-3\end{matrix}\right.\)

vậy tập nghiệm của pt là S={1; -3}

3, 2(x+2)²-x³-8=0

\(\Leftrightarrow\)2(x+2)²-(x³+8)=0

\(\Leftrightarrow\)(x+2)[2(x+2)-(x²-2x+4)]=0

\(\Leftrightarrow\)-x(x+2)(4-x)=0

\(\Leftrightarrow\)\(\left[{}\begin{matrix}\text{-x=0}\\\text{x}+2=0\\4-x=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left[{}\begin{matrix}\text{x=0}\\\text{x}=-2\\x=4\end{matrix}\right.\)

vậy tập nghiệm của pt là S={0; -2; 4}

4, x³-3x²+3x-1=(x-1)(x+1)

\(\Leftrightarrow\)(x³-1)+(-3x²+3x)=(x-1)(x+1)

\(\Leftrightarrow\)(x-1)(x²+x+1)-3x(x-1)=(x-1)(x+1)

\(\Leftrightarrow\)(x-1)(x²-2x+1)-(x-1)(x+1)=0

\(\Leftrightarrow\)(x-1)(x²-3x)=0

\(\Leftrightarrow\)x(x-1)(x-3)=0

\(\Leftrightarrow\)\(\left[{}\begin{matrix}\text{x=0}\\\text{x}-1=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left[{}\begin{matrix}\text{x=0}\\\text{x}=1\\x=3\end{matrix}\right.\)

vậy tập nghiệm của pt là S={0;1;3}