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13 tháng 7 2016

a) \(\left(\sqrt{ab}+2\sqrt{\frac{b}{a}}-\sqrt{\frac{a}{b}}+\frac{1}{\sqrt{ab}}\right).\sqrt{ab}\) (ĐK : \(\hept{\begin{cases}a>0\\b>0\end{cases}}\)hoặc \(\hept{\begin{cases}a< 0\\b< 0\end{cases}}\))

\(=ab+2b-a+1\)

b) \(\left(-\frac{am}{b}\sqrt{\frac{n}{m}}-\frac{ab}{n}.\sqrt{mn}+\frac{a^2}{b^2}.\sqrt{\frac{m}{n}}\right)\left(a^2b^2.\sqrt{\frac{n}{m}}\right)\) (ĐK bạn tự xét nhé ^^)

\(=\left(-\frac{a\sqrt{mn}}{b}-\frac{ab\sqrt{m}}{\sqrt{n}}+\frac{a^2}{b^2}.\sqrt{\frac{m}{n}}\right)\left(a^2b^2.\sqrt{\frac{n}{m}}\right)\)

\(=a^2b^2\left(\frac{-an}{b}-ab+\frac{a^2}{b^2}\right)=-a^3bn-a^3b^3+a^4=a^3\left(a-bn-b^3\right)\)

19 tháng 8 2019

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

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

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

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

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

\(=\frac{4\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)

19 tháng 8 2019

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

\(=\left(\frac{\sqrt{a}^3+\sqrt{b}^3}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right)^2\)

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

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

\(=\left(\sqrt{a}-\sqrt{b}\right)^2.\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)^2}=1\)

4 tháng 10 2015

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

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

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

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

\(=\frac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)

\(=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}-\sqrt{b}}=1\)

tick cho mình nha

4 tháng 10 2015

trục căn ở mẫu là đc :D

4 tháng 10 2016

ui khó quá mình không biết đâu