a) \(\sqrt{\dfrac{a}{b}}+\sqrt{ab}+\dfrac{a}{b}\sqrt{\dfrac{a}{b}}\) với \(a>0;b>0\)
b) \(\sqrt{\dfrac{m}{1-2x+x^2}}.\sqrt{\dfrac{4m-8mx+4mx^2}{81}}\) vớ \(m>0;x\ne1\)
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30 tháng 8 2022
a: \(=\dfrac{\sqrt{a}-1}{\sqrt{a}\left(a-\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{1}\)
\(=a-1\)
b: \(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\left(\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}+\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\dfrac{\sqrt{ab}+b+\sqrt{ab}-b}{\sqrt{a}\left(a-b\right)}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{1}{\sqrt{a}}\)
c: \(=\dfrac{a\sqrt{b}+b}{a-b}\cdot\sqrt{\dfrac{ab+b^2-2b\sqrt{ab}}{a^2+2a\sqrt{b}+b}}\cdot\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\sqrt{\dfrac{b\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(a+\sqrt{b}\right)^2}}\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{a+\sqrt{b}}=b\)
13 tháng 7 2018
b)CM: \(ab\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\)
\(VT=ab\sqrt{\dfrac{a^2b^2+1}{\left(ab\right)^2}}-\sqrt{a^2b^2+1}\)
\(VT=ab\dfrac{\sqrt{a^2b^2+1}}{ab}-\sqrt{a^2b^2+1}\)
\(VT=\sqrt{a^2b^2+1}-\sqrt{a^2b^2+1}\)
\(VT=0=VP\)
AP
1
13 tháng 8 2018
\(\sqrt{\dfrac{a}{b}}+\sqrt{ab}+\dfrac{a}{b}\sqrt{\dfrac{b}{a}}=\dfrac{\sqrt{a}}{\sqrt{b}}+\dfrac{a}{b}.\dfrac{\sqrt{b}}{\sqrt{a}}+\sqrt{ab}=\dfrac{2\sqrt{a}}{\sqrt{b}}+\sqrt{ab}=\dfrac{2\sqrt{a}+b\sqrt{a}}{\sqrt{b}}=\dfrac{\sqrt{a}\left(2+b\right)}{\sqrt{b}}\)
9 tháng 7 2022
a: \(\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\right)\)
\(=\left(a-\sqrt{ab}+b-\sqrt{ab}\right)\cdot\dfrac{1}{\sqrt{a}-\sqrt{b}}\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}=\sqrt{a}-\sqrt{b}\)
b: \(VT=\dfrac{\sqrt{2}\left(2+\sqrt{2}\right)}{2+\sqrt{3}+1}+\dfrac{\sqrt{2}\left(2-\sqrt{2}\right)}{2-\left(\sqrt{3}-1\right)}\)
\(=\dfrac{\sqrt{2}\left(2+\sqrt{2}\right)}{3+\sqrt{3}}+\dfrac{\sqrt{2}\left(2-\sqrt{2}\right)}{3-\sqrt{3}}\)
\(=\dfrac{2\left(\sqrt{2}+1\right)\left(\sqrt{3}-1\right)+2\left(\sqrt{2}-1\right)\left(\sqrt{3}+1\right)}{\sqrt{3}\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=\dfrac{2\left(\sqrt{6}-\sqrt{2}+\sqrt{3}-1+\sqrt{6}+\sqrt{2}-\sqrt{3}-1\right)}{\sqrt{3}\cdot2}\)
\(=\dfrac{2\left(2\sqrt{6}-2\right)}{2\sqrt{3}}=\dfrac{2\sqrt{6}-2}{\sqrt{3}}\)
1 tháng 10 2018
Ta có VT =\(\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{a-b}\)
=\(\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}-\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}-\dfrac{2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\) =\(\dfrac{a+\sqrt{ab}-\sqrt{ab}+b-2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
=\(\dfrac{a-b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
=\(\dfrac{a-b}{a-b}=1=VP\)
17 tháng 11 2021
\(=\dfrac{\sqrt{ab}}{b}+\sqrt{\dfrac{a^2b}{b^2a}}=\dfrac{\sqrt{ab}}{b}+\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{ab}}{b}+\dfrac{\sqrt{ab}}{b}=\dfrac{2\sqrt{ab}}{b}\left(B\right)\)
4 tháng 9 2022
1: \(Q=\dfrac{ab\left(a-b\right)}{ab}\cdot\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\left(\sqrt{a}+\sqrt{b}\right)^2=a+2\sqrt{ab}+b\)
2: \(=\dfrac{-1+\sqrt{5}-\sqrt{5}+\sqrt{9}-...-\sqrt{2001}+\sqrt{2005}}{4}\)
\(=\dfrac{\sqrt{2005}-1}{4}\)
20 tháng 6 2018
Câu a
\(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}\)
\(=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}\)
\(=\left(\sqrt{a}+\sqrt{b}\right):\dfrac{1}{\sqrt{a}-\sqrt{b}}\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{1}\)
\(=a-b\)
a) \(\sqrt{\dfrac{a}{b}}+\sqrt{ab}+\dfrac{a}{b}\sqrt{\dfrac{a}{b}}\) với a>0 và b>0
b) \(\sqrt{\dfrac{m}{1-2x+x^2}}.\sqrt{\dfrac{4m-8mx+4mx^2}{81}}=\sqrt{\dfrac{m}{1-2x+x^2}}.\sqrt{\dfrac{4m\left(2-2x+x^2\right)}{81}}\)
\(=\sqrt{\dfrac{4m^2\left(1-2x+x^2\right)}{81\left(1-2x+x^2\right)}}=\sqrt{\dfrac{4m^2}{81}}=\sqrt{\dfrac{2m}{9}}\)