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QD
22 tháng 4 2017
a) \(\sqrt{\dfrac{a}{b}}+\sqrt{ab}+\dfrac{a}{b}\sqrt{\dfrac{a}{b}}\) với a>0 và b>0
b) \(\sqrt{\dfrac{m}{1-2x+x^2}}.\sqrt{\dfrac{4m-8mx+4mx^2}{81}}=\sqrt{\dfrac{m}{1-2x+x^2}}.\sqrt{\dfrac{4m\left(2-2x+x^2\right)}{81}}\)
\(=\sqrt{\dfrac{4m^2\left(1-2x+x^2\right)}{81\left(1-2x+x^2\right)}}=\sqrt{\dfrac{4m^2}{81}}=\sqrt{\dfrac{2m}{9}}\)
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: \(=\dfrac{a\sqrt{b}+b}{a-b}\cdot\sqrt{\dfrac{ab+b^2-2b\sqrt{ab}}{a^2+2a\sqrt{b}+b}}\cdot\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\sqrt{\dfrac{b\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(a+\sqrt{b}\right)^2}}\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{a+\sqrt{b}}=b\)
5 tháng 9 2022
a: \(=6\sqrt{a}+\dfrac{1}{3}\sqrt{a}-3\sqrt{a}+\sqrt{7}=\dfrac{10}{3}\sqrt{a}+\sqrt{7}\)
b: \(=5a\cdot5b\sqrt{ab}+\sqrt{3}\cdot2\sqrt{3}\cdot ab\sqrt{ab}+9ab\cdot3\sqrt{ab}-5b\cdot9a\sqrt{ab}\)
\(=25ab\sqrt{ab}+12ab\sqrt{ab}+27ab\sqrt{ab}-45ab\sqrt{ab}\)
\(=19ab\sqrt{ab}\)
c: \(=\dfrac{\sqrt{ab}}{b}+\sqrt{ab}-\dfrac{a}{b}\cdot\dfrac{\sqrt{b}}{\sqrt{a}}\)
\(=\sqrt{ab}\left(\dfrac{1}{b}+1\right)-\dfrac{\sqrt{a}}{\sqrt{b}}\)
\(=\sqrt{ab}\)
d: \(=11\sqrt{5a}-5\sqrt{5a}+2\sqrt{5a}-12\sqrt{5a}+9\sqrt{a}\)
\(=-4\sqrt{5a}+9\sqrt{a}\)
20 tháng 6 2018
Câu a
\(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}\)
\(=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}\)
\(=\left(\sqrt{a}+\sqrt{b}\right):\dfrac{1}{\sqrt{a}-\sqrt{b}}\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{1}\)
\(=a-b\)
29 tháng 7 2022
a: \(=4\left|a-3\right|=4\left(a-3\right)=4a-12\)
b: \(=9\cdot\left|a-9\right|=9\left(9-a\right)=81-9a\)
c: \(a^3b^6\cdot\sqrt{\dfrac{3}{a^6b^4}}=a^3b^6\cdot\dfrac{\sqrt{3}}{-a^3b^2}=-b^4\sqrt{3}\)
d: \(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{a-b}\)
\(=\dfrac{a+\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}\)
13 tháng 7 2018
b)CM: \(ab\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\)
\(VT=ab\sqrt{\dfrac{a^2b^2+1}{\left(ab\right)^2}}-\sqrt{a^2b^2+1}\)
\(VT=ab\dfrac{\sqrt{a^2b^2+1}}{ab}-\sqrt{a^2b^2+1}\)
\(VT=\sqrt{a^2b^2+1}-\sqrt{a^2b^2+1}\)
\(VT=0=VP\)
26 tháng 5 2018
Áp dụng bđt AM-GM: \(\dfrac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\le\dfrac{2\sqrt{ab}}{2\sqrt{\sqrt{ab}}}=\sqrt{\sqrt{ab}}\)
NT
1
3 tháng 12 2018
\(VT=\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}.\left(\sqrt{a}-\sqrt{b}\right)=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)=a-b\)
4 tháng 7 2017
VT= \(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}=\dfrac{\sqrt{a^3}+\sqrt{b^3}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a^2}-\sqrt{ab}+\sqrt{b^2}\right)}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\)
\(=\sqrt{a^2}-\sqrt{ab}+\sqrt{b^2}-\sqrt{ab}\)\(=\sqrt{a^2}-2\sqrt{ab}+\sqrt{b^2}=\left(\sqrt{a}-\sqrt{b}\right)^2\)=VP
=> đpcm.
\(=\dfrac{\sqrt{a}\left(\sqrt{ab}+1\right)-\sqrt{b}\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)
\(=\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{ab}-1}\)