Bài 1:
\(a,=2\sqrt{3a}-5\sqrt{3a}+\dfrac{1}{2}\cdot4\sqrt{3a}=-3\sqrt{3a}+2\sqrt{3a}=-\sqrt{3a}\\ b,=\dfrac{a+b}{b^2}\cdot\dfrac{\left|a\right|b^2}{\left|a+b\right|}=\dfrac{a+b}{b^2}\cdot\dfrac{\left|a\right|b^2}{a+b}=\left|a\right|\\ c,=5\sqrt{a}-20\left|a\right|b\sqrt{a}+20a\left|b\right|\sqrt{a}-6\sqrt{a}\\ =\left(5\sqrt{a}-6\sqrt{a}\right)-\left(20ab\sqrt{a}-20ab\sqrt{a}\right)\\ =-\sqrt{a}\\ d,=\dfrac{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)}{x+\sqrt{3}}=x-\sqrt{3}\\ e,=\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}=1+\sqrt{a}+a\)
Bài 2:
\(a,B=7\sqrt{x+2}-4\sqrt{x+2}-2\sqrt{x+2}=\sqrt{x+2}\\ b,ĐK:x\ge-2\\ PT\Leftrightarrow\sqrt{x+2}=20\Leftrightarrow x+2=400\Leftrightarrow x=398\left(tm\right)\)
Bài 3:
\(a,M=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ b,M=\dfrac{\sqrt{a}-1}{\sqrt{a}}=1-\dfrac{1}{\sqrt{a}}< 1\left(\dfrac{1}{\sqrt{a}}>0\right)\)
