\(\frac{\left(\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{1+\sqrt{ab}}\right)}{\left(1+\frac{a+b+2ab}{1-ab}\right)}\)
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SL
5 tháng 6 2017
mình thật sự muốn lm cho bạn nhưng nhìu việc quá,,,chỗ nào ko làm đc bạn hỏi mình,,,,mình làm cho
27 tháng 6 2020
ĐK: ab khác 1; a,b \(\ge\)0
\(B=\left(\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{1+\sqrt{ab}}\right):\left(1+\frac{a+b+2ab}{1-ab}\right)\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(1+\sqrt{ab}\right)+\left(\sqrt{a}-\sqrt{b}\right)\left(1-\sqrt{ab}\right)}{\left(1-\sqrt{ab}\right)\left(1+\sqrt{ab}\right)}:\frac{1-ab+a+b+2ab}{1-ab}\)
\(=\frac{2\sqrt{a}+2\sqrt{b}\sqrt{ab}}{1-ab}:\frac{1+ab+a+b}{1-ab}\)
\(=\frac{2\sqrt{a}\left(1+b\right)}{1-ab}:\frac{\left(1+b\right)\left(1+a\right)}{1-ab}\)
\(=\frac{2\sqrt{a}}{1+a}\)
T
20 tháng 8 2017
\(A=\left(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\right)\left[\left(\frac{1}{\sqrt{a}-\sqrt{b}}-\frac{3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right):\frac{a-b}{a+\sqrt{ab}+b}\right]\)
\(A=\left[\frac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\right].\left[\frac{a+b+\sqrt{ab}-3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}.\frac{a+\sqrt{ab}+b}{a-b}\right]\)
\(A=\left[\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\right].\left[\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right]\)
\(A=\frac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}.\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{1}{a-\sqrt{ab}+b}\)
Điều kiện : a, b\(\ge0\)
\(\frac{\left(\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{1+\sqrt{ab}}\right)}{\left(1+\frac{a+b+2ab}{1-ab}\right)}\)
Tử số: \(\left(\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{1+\sqrt{ab}}\right)=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(1+\sqrt{ab}\right)+\left(\sqrt{a}-\sqrt{b}\right)\left(1-\sqrt{ab}\right)}{\left(1-\sqrt{ab}\right)\left(1+\sqrt{ab}\right)}\)
\(=\frac{\sqrt{a}+\sqrt{b}+a\sqrt{b}+b\sqrt{a}+\sqrt{a}-\sqrt{b}-a\sqrt{b}+b\sqrt{a}}{1-ab}\)
\(=\frac{2\sqrt{a}+2\sqrt{a}b}{1-ab}=\frac{2\sqrt{a}\left(1+b\right)}{1-ab}\)
Mẫu số: \(1+\frac{a+b+2ab}{1-ab}=\frac{1-ab+a+b+2ab}{1-ab}=\frac{1+ab+a+b}{1-ab}=\frac{\left(a+b\right)\left(b+1\right)}{1-ab}\)
Do đó:
\(\frac{\left(\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{1+\sqrt{ab}}\right)}{\left(1+\frac{a+b+2ab}{1-ab}\right)}=\frac{\frac{2\sqrt{a}\left(1+b\right)}{1-ab}}{\frac{\left(a+1\right)\left(b+1\right)}{1-ab}}=\frac{2\sqrt{a}}{a+1}\)