Question

M=\(\frac{1}{3-\sqrt{x}}+\frac{\sqrt{x}}{3+\sqrt{x}}-\frac{x+9}{x-9}\)

a) rút gọn N

b)tìm x để M>1

Answer 1

a) Ta có: \(M=\frac{1}{3-\sqrt{x}}+\frac{\sqrt{x}}{3+\sqrt{x}}-\frac{x+9}{x-9}\)

\(=\frac{3+\sqrt{x}}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}+\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\)

\(=\frac{3+\sqrt{x}+3\sqrt{x}-x+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\)

\(=\frac{4\sqrt{x}+12}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\)

\(=\frac{4\left(\sqrt{x}+3\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\)

\(=\frac{4}{3-\sqrt{x}}\)

Answer 2

Lời giải:

ĐKXĐ: $x\geq 0; x\neq 9$

a) Bạn Thịnh đã làm rồi.

b)

$M=\frac{4}{3-\sqrt{x}}>1\Leftrightarrow \frac{4}{3-\sqrt{x}}-1>0$

$\Leftrightarrow \frac{1+\sqrt{x}}{3-\sqrt{x}}>0$

$\Leftrightarrow 3-\sqrt{x}>0\Leftrightarrow \sqrt{x}< 3\Leftrightarrow 0\leq x< 9$

Kết hợp với ĐKXĐ ta suy ra $0\leq x< 9$ là đáp án cuối cùng.