Question

Answer 1

a) \(=\dfrac{\left(\sqrt{x}-2+\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}-2}{\sqrt{x}}=\dfrac{2}{\sqrt{x}+2}\)
b) Vì \(\sqrt{x}>0=>3\sqrt{x}+6>6=>\left[{}\begin{matrix}3\sqrt{x}+6=7\\3\sqrt{x}+6=14\end{matrix}\right.< =>\left[{}\begin{matrix}x=\dfrac{1}{9}\\x=\dfrac{64}{9}\end{matrix}\right.\)