a)
\(\sqrt{\dfrac{27a^4}{48a^2}}=\sqrt{\dfrac{9a^2}{16}}=\sqrt{\left(\dfrac{3a}{4}\right)^2}=\dfrac{3a}{4}\)
b)
\(\dfrac{\sqrt{9x^2-25}}{\sqrt{3x+5}}=\dfrac{\sqrt{\left(3x-5\right)\left(3x+5\right)}}{\sqrt{3x+5}}=\sqrt{3x-5}\)
c)
\(\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2\\ =\left(3-\sqrt{5}\right)+2.\sqrt{3-\sqrt{5}}.\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\\ =2.\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}+6\\ =2.\sqrt{9-5}+6\\ =10\\ \Rightarrow\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}=\sqrt{10}\)
d) KO khó!!
a) \(\sqrt{\dfrac{27a^2}{48a^4}}=\sqrt{\dfrac{9}{16a^2}}=\dfrac{3}{4a}\)
b) \(\dfrac{\sqrt{9x^2-25}}{\sqrt{3x+5}}=2\Leftrightarrow\dfrac{\sqrt{\left(3x-5\right)\left(3x+5\right)}}{\sqrt{3x+5}}=2\Leftrightarrow\sqrt{3x-5}=2\Leftrightarrow3x-5=4\Leftrightarrow x=3\)c) \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}=\dfrac{\sqrt{2}\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)}{\sqrt{2}}=\dfrac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\dfrac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)d) \(\dfrac{a\sqrt{a}+b\sqrt{b}-a\sqrt{b}-b\sqrt{a}}{a+b-2\sqrt{ab}}=\dfrac{a\left(\sqrt{a}-\sqrt{b}\right)-b\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)^2}=\dfrac{\left(a-b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)^2}=\sqrt{a}+\sqrt{b}\)
