Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\frac{\sqrt{a}}{\sqrt{a}-1}+\frac{\sqrt{a}}{a-1}\right):\left(\frac{2}{a}-\frac{2-a}{a\sqrt{a}+a}\right)\)
=\(\frac{\sqrt{a}\left(\sqrt{a}+1\right)+\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}:\frac{2\left(\sqrt{a}+1\right)-2+a}{a\left(\sqrt{a}+1\right)}\)
= \(\frac{\sqrt{a}\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}.\frac{a\left(\sqrt{a+1}\right)}{\sqrt{a}\left(2+\sqrt{a}\right)}\)
=\(\frac{a}{\sqrt{a}-1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(VT=\frac{\sqrt{a}+\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}-\sqrt{b}}{2\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2b}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)\(=\frac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{4\sqrt{ab}+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}=VP\)(ĐPCM)
2) \(VT=\text{[}\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a+b-\sqrt{ab}\right)}{\left(\sqrt{a}+\sqrt{b}\right)}-\sqrt{ab}\text{]}.\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}\)
\(=\frac{\left(a+b-\sqrt{ab}-\sqrt{ab}\right)\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}\)\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}=\frac{\left(a-b\right)^2}{\left(a-b\right)^2}=1=VP\)(ĐPCM)
4) \(VT=\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a=VP\)(ĐPCM)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=(2+\(\frac{3+\sqrt{3}}{\sqrt{3}+1}\)) . (2-\(\frac{3-\sqrt{3}}{\sqrt{3}-3}\))
=(\(2+\frac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}\)) . (\(2-\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\))
=(\(2+\sqrt{3}\)) . (\(2-\sqrt{3}\))
=22-(\(\sqrt{3}\))2=4-3=1
B=(\(\frac{\sqrt{b}}{a-\sqrt{ab}}-\frac{\sqrt{a}}{\sqrt{ab}-b}\)) . (\(a\sqrt{b}-b\sqrt{a}\))
=(\(\frac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}\)) . (\(a\sqrt{b}-b\sqrt{a}\))
=(\(\frac{\sqrt{b}.\sqrt{b}}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}.\sqrt{a}}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\)). (a\(\sqrt{b}-b\sqrt{a}\))
=\(\frac{b-a}{\sqrt{ab}.\left(\sqrt{a}-\sqrt{b}\right)}.\left(\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\right)\)
=b-a
Ta có: \(A=\left(2+\frac{3+\sqrt{3}}{\sqrt{3}+1}\right)\cdot\left(2-\frac{3-\sqrt{3}}{\sqrt{3}-1}\right)\)
\(=\frac{2\left(\sqrt{3}+1\right)+3+\sqrt{3}}{\sqrt{3}+1}\cdot\frac{2\left(\sqrt{3}-1\right)-3+\sqrt{3}}{\sqrt{3}-1}\)
\(=\frac{2\sqrt{3}+2+3+\sqrt{3}}{\sqrt{3}+1}\cdot\frac{2\sqrt{3}-2-3+\sqrt{3}}{\sqrt{3}-1}\)
\(=\frac{3\sqrt{3}+5}{\sqrt{3}+1}\cdot\frac{3\sqrt{3}-5}{\sqrt{3}-1}\)
\(=\frac{2}{2}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ:...
\(V=\left(\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\frac{x-1-2\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\)
\(=\left(\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\frac{x-2\sqrt{x}-3}{\sqrt{x}+1}\right)=\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+1\right)}=\frac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(W=\left(\frac{\sqrt{a}-1}{a+\sqrt{a}+1}-\frac{a-3\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}-\frac{1}{\sqrt{a}-1}\right).\left(\frac{1-\sqrt{a}}{a+1}\right)\)
\(=\left(\frac{\left(\sqrt{a}-1\right)^2-a+3\sqrt{a}-1-\left(a+\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\right)\left(\frac{1-\sqrt{a}}{a+1}\right)\)
\(=\left(\frac{-\left(a+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\right)\left(\frac{-\left(\sqrt{a}-1\right)}{a+1}\right)=\frac{1}{a+\sqrt{a}+1}\)