Đáp án:
Giải thích các bước giải:
a) Ta có: $M=\sqrt{a}.\sqrt{b}=\sqrt{ab}$
$⇒M=\sqrt{2.8}=\sqrt{16}=4$
b) Ta có: `N=\sqrt{c^2}-\frac{1}{c}=|c|-\frac{1}{c}`
`⇒N=|\sqrt{5}-2|-\frac{1}{\sqrt{5}-2}`
`=\sqrt{5}-2-\frac{1}{\sqrt{5}-2}`
`=\frac{(\sqrt{5}-2)^2-1}{\sqrt{5}-2}`
`=\frac{9-4\sqrt{5}-1}{\sqrt{5}-2}`
`=\frac{8-4\sqrt{5}}{\sqrt{5}-2}`
`=\frac{-4(\sqrt{5}-2)}{\sqrt{5}-2}`
`=-4`
c) $2x^2+x(2c-\sqrt{a})-c\sqrt{2}=0$
$⇔2x^2+2cx-x\sqrt{a}-c\sqrt{2}=0$
$⇔2x^2+2cx-x\sqrt{2}-c\sqrt{2}=0$
$⇔2x(x+c)-\sqrt{2}(x+c)=0$
$⇔(2x-\sqrt{2})(x+c)=0$
$⇔\left[ \begin{array}{l}2x-\sqrt{2}=0\\x+c=0\end{array} \right.$
$⇔\left[ \begin{array}{l}2x=\sqrt{2}\\x=-c\end{array} \right.$
$⇔\left[ \begin{array}{l}x=\frac{\sqrt{2}}{2}\\x=-(\sqrt{5}-2)=2-\sqrt{5}\end{array} \right.$
