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![](https://rs.olm.vn/images/avt/0.png?1311)
b) \(\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
\(=\left(\dfrac{\sqrt{7}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right).\left(\sqrt{7}-\sqrt{5}\right)\)
\(=\left(-\sqrt{7}-\sqrt{5}\right).\left(\sqrt{7}-\sqrt{5}\right)=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
\(=-\left(7-5\right)=-2\)
c) \(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}.\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)=a-b\)
d) \(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\left(1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)
b) \(\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
\(=\left(\dfrac{\sqrt{7}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right).\left(\sqrt{7}-\sqrt{5}\right)\)
\(=\left(-\sqrt{7}+\left(-\sqrt{5}\right)\right).\left(\sqrt{7}-\sqrt{5}\right)=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a 3 b 3 = a 3 3 . b 3 3 = a b 3
Vế trái bằng vế phải nên đẳng thức được chứng minh.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(-\sqrt{121a^2}+3\sqrt{36a^2}+\sqrt{49a^2}=-11\left|a\right|+18\left|a\right|+7\left|a\right|=11a-18a-7a=-14a\left(đpcm\right)\left(do.a< 0\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a b 2 3 = a b b 3 3 = 1 b a b 3 b ≠ 0
Vế trái bằng vế phải nên đẳng thức được chứng minh.
![](https://rs.olm.vn/images/avt/0.png?1311)
= (-√7 - √5)(√7 - √5)
= -(√7 + √5)(√7 - √5)
= -(7 - 5) = -2 = VP (đpcm)