tth

\(A=\left(\frac{1}{\sqrt{a}-3}+\frac{1}{\sqrt{a}+3}\right)\left(1-\frac{3}{\sqrt{a}}\right)\) \(đk:a>0;a\ne9\)

\(=\frac{\sqrt{a}+3+\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}.\frac{\sqrt{a}-3}{\sqrt{a}}\)

\(=\frac{2\sqrt{a}}{\sqrt{a}\left(\sqrt{a}+3\right)}\)

\(=\frac{2}{\sqrt{a}+3}\)

\(đk:a>0;a\ne9\)

\(A>\frac{1}{2}=>\frac{2}{\sqrt{a}+3}>\frac{1}{2}\)

\(=>4>\sqrt{a}+3\)

\(< =>\sqrt{a}>1\)

\(< =>a=1\)

ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\b>0\\a\ne b\end{matrix}\right.\)

\(A=\frac{a+b+2\sqrt{ab}-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)

\(A=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\left(\sqrt{a}+\sqrt{b}\right)\)

\(A=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)

Lời giải:
a)

ĐK: \(a,b>0; a\neq b\)

b)

\(A=\frac{(\sqrt{a}+\sqrt{b})^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}=\frac{a+2\sqrt{ab}+b-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{ab}(\sqrt{a}+\sqrt{b})}{\sqrt{ab}}\)

\(=\frac{a-2\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}-(\sqrt{a}+\sqrt{b})\)

\(=\frac{(\sqrt{a}-\sqrt{b})^2}{\sqrt{a}-\sqrt{b}}-(\sqrt{a}+\sqrt{b})=(\sqrt{a}-\sqrt{b})-(\sqrt{a}+\sqrt{b})=-2\sqrt{b}\)

\(Q=\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}=2\sqrt{b}\)

DK: \(a,b\ge0\)do \(Q=2\sqrt{b}\)nên Q ko phụ thuộc vào giá trị của a

a) \(\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}}=\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}=\sqrt{ab}\)

b) Giống câu a ?

c) \(\left(\sqrt{ab}-\sqrt{\frac{a}{b}}+\frac{1}{a}\sqrt{4ab}+\frac{1}{b}\sqrt{\frac{b}{a}}\right):\left(1+\frac{2}{a}-\frac{1}{b}+\frac{1}{ab}\right)\)

\(=\left(\sqrt{ab}-\sqrt{\frac{a}{b}}+\sqrt{\frac{4b}{a}}+\sqrt{\frac{1}{ab}}\right):\left(\frac{ab+2b-a+1}{ab}\right)\)

\(=\frac{ab-a+2b+1}{\sqrt{ab}}\cdot\frac{ab}{ab+2b-a+1}\)

\(=\sqrt{ab}\)