Đáp án:
Giải thích các bước giải:
a) ĐK: `x \ge 0, x \ne 9`
`B=\frac{1}{\sqrt{x}-3}+\frac{1}{\sqrt{x}+3}`
`B=\frac{\sqrt{x}+3}{(\sqrt{x}-3)(\sqrt{x}+3)}+\frac{\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}`
`B=\frac{2\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}`
b) `A=(\sqrt{8}-\sqrt{12})(\sqrt{2}+\sqrt{3})`
`A=-2`
Để `A=B`
`\frac{2\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}=-2`
`⇔ -2(x-9)=2\sqrt{x}`
`⇔ 2x+2\sqrt{x}-18=0`
`⇔` \(\left[ \begin{array}{l}x=\dfrac{-1+\sqrt{37}}{2}\ (TM)\\x=\dfrac{-1-\sqrt{37}}{2}\ (L)\end{array} \right.\)
Vậy \(x=\dfrac{-1+\sqrt{37}}{2}\)
