rút gọn các biểu thức:
a,\(6\sqrt{a}+\dfrac{2}{3}\sqrt{\dfrac{a}{4}}-a\sqrt{\dfrac{9}{a}}+\sqrt{7}vớia>0\)
b,\(5a\sqrt{25ab^3}\sqrt{3}\sqrt{12a^3b^3}+9ab\sqrt{9ab}-5b\sqrt{81a^3b}vớia,b>0\)
c,\(\sqrt{\dfrac{a}{b}}+\sqrt{ab}-\dfrac{a}{b}\sqrt{\dfrac{b}{a}}vớia,b>0\)
d,\(11\sqrt{5a}-\sqrt{125a}+\sqrt{20a}-4\sqrt{45a}+9\sqrt{a}vớia>0\)
Bạn làm đc bài này chưa chỉ mình với
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: \(=11\sqrt{5a}-5\sqrt{5a}+2\sqrt{5a}-12\sqrt{5a}+9\sqrt{a}\)
\(=-4\sqrt{5a}+9\sqrt{a}\)