Đáp án:
Giải thích các bước giải:
a,`(x-\sqrt{2})+3(x^2-2)=0`
`⇔ (x-\sqrt{2})+3(x-\sqrt{2})(x+\sqrt{2})=0`
`⇔ (x-\sqrt{2})[1+3(x+\sqrt{2})]=0`
`⇔` \(\left[ \begin{array}{l}x-\sqrt{2}=0\\1+3(x+\sqrt{2})=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\sqrt{2}\\x=\dfrac{-1-3\sqrt{2}}{3}\end{array} \right.\)
Vậy `S={\sqrt{2};\frac{-1-3\sqrt{2}}{3}}`
b, `x^2-5=(2x-\sqrt{5})(x+\sqrt{5})`
`⇔ (x-\sqrt{5})(x+\sqrt{5})=(2x-\sqrt{5})(x+\sqrt{5})`
`⇔ (x-\sqrt{5})(x+\sqrt{5})-(2x-\sqrt{5})(x+\sqrt{5})=0`
`⇔ (x+\sqrt{5})(x-\sqrt{5}-2x+\sqrt{5})=0`
`⇔ (x+\sqrt{5}).(-x)=0`
`⇔` \(\left[ \begin{array}{l}x=0\\x=-\sqrt{5}\end{array} \right.\)
Vậy `S={0;-\sqrt{5}}`
