\(\sqrt{\dfrac{a^3}{a}}=\sqrt{a^2}=\left|a\right|=-a\)

Với a > 1, ta có

\(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}=\dfrac{\sqrt{a\left(\sqrt{a-1}\right)}}{1-\sqrt{a}}\dfrac{-\sqrt{a\left(1-\sqrt{a}\right)}}{1-\sqrt{a}}=\sqrt{a}\)

\(1,=0,9\left|x\right|\\ 2,Sửa:\dfrac{\sqrt{63y^3}}{\sqrt{7y}}=\sqrt{\dfrac{63y^3}{7y}}=\sqrt{9y^2}=3\left|y\right|=-3y\)

Rút gọn biểu thức 

c

B

với a < 0

với a<0

\(-\dfrac{\sqrt{x^2}}{x}=\dfrac{-\left|x\right|}{x}=\dfrac{-x}{x}=-1\)(Vì x>0 nên |x|=x)

D

\(B=\dfrac{a+3\sqrt{a}-3\sqrt{a}+9-a+2}{a-9}=\dfrac{11}{a-9}\)

Câu 2: 

Ta có: \(M=\left(\dfrac{a+\sqrt{a}}{\sqrt{a}+1}+1\right)\left(1+\dfrac{a-\sqrt{a}}{1-\sqrt{a}}\right)\)

\(=\left(\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}+1\right)\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)

\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\)

\(=1-a\)

Câu 1: 

Ta có: \(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)

\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1}{\sqrt{a}+1}\right)^2\)

\(=\left(\sqrt{a}+1\right)^2\cdot\dfrac{1}{\left(\sqrt{a}+1\right)^2}\)

\(=1\)