Đáp án+Giải thích các bước giải:
`b)`
`2x^3-4x=0`
`⇔2x(x^2-2)=0`
\(⇔\left[ \begin{array}{l}2x=0\\x^2-2=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=0\\x^2=2\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=0\\x=±\sqrt2\end{array} \right.\)
Vậy `x=0` hoặc `x=\sqrt2` hoặc `x=-\sqrt2`
`c)`
`x^2-4x-1=0`
`⇔x^2-4x+4-5=0`
`⇔x^2-2.x.2+2^2-5=0`
`⇔(x-2)^2-(\sqrt5)^2=0`
`⇔(x-2)^2=(\sqrt5)^2`
`⇔x-2=±\sqrt5`
`⇔x=2±\sqrt5`
Vậy `x=2+\sqrt5` hoặc `x=2-\sqrt5`
`d)`
`2x^3+3x^2+2x-2=0`
`⇔2x^3-x^2+4x^2-2x+4x-2=0`
`⇔x^2(2x-1)+2x(2x-1)+2(2x-1)=0`
`⇔(2x-1)(x^2+2x+2)=0`
TH 1:
`2x-1=0`
`⇔2x=1`
`⇔x=1/2`
TH 2:
`x^2+2x+2=0`
`⇔x^2+2x+1+1=0`
`⇔(x+1)^2+1=0`
`⇔(x+1)^2=-1` (Vô lí vì `(x+1)^2≥0∀x`)
Vậy `x=1/2`
