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14 tháng 8 2019

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

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

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

\(=\sqrt{ab}\)

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\(\ge0\)

12 tháng 5 2019

\(A=\left(\frac{\sqrt{a}}{\sqrt{ab}-b}+\frac{\sqrt{b}}{\sqrt{ab}-a}\right):\frac{\sqrt{a}+\sqrt{b}}{a\sqrt{b}-b\sqrt{a}}=\left[\frac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}+\frac{\sqrt{b}}{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}\right].\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{a}+\sqrt{b}}=\left[\frac{a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{b}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\right].\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}=\frac{\left(a-b\right)\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\frac{a-b}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}=\sqrt{a}-\sqrt{b}\)

4 tháng 1 2020

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

23 tháng 7 2018

a)  ĐK:  a > 0;  b > 0

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

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

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

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

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

b)  \(A=1\)\(\Rightarrow\)\(2\sqrt{b}-b=1\)

                    \(\Leftrightarrow\)\(b-2\sqrt{b}+1=0\)

                   \(\Leftrightarrow\) \(\left(\sqrt{b}-1\right)^2=0\)

                   \(\Leftrightarrow\)\(\sqrt{b}-1=0\)

                   \(\Leftrightarrow\)\(\sqrt{b}=1\)

                   \(\Leftrightarrow\)\(b=1\)   (t/m ĐKXĐ)

Vậy  b=1

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

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

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

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

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

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