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

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

\(=2+\dfrac{2a+2}{\sqrt{a}}\)

\(=\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}\)

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

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

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

\(=1\)

26 tháng 7 2021

A=\(\left[\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}+1\right)}{\left(a-1\right)\left(\sqrt{a}+2\right)}-\dfrac{\left(a+\sqrt{a}\right)}{\left(a-1\right)}\right]\)::::::::\(\left(\dfrac{\left(\sqrt{a}-1+\sqrt{a}+1\right)}{a-1}\right)\)

=\(\left[\dfrac{1}{\sqrt{a}-1}\right]:\left(\dfrac{2\sqrt{a}}{a-1}\right)\)=\(\dfrac{\sqrt{a}-1}{2\sqrt{a}}\)

=\(\dfrac{a^2+a\sqrt{a}+11a+6}{2\sqrt{a}\left(\sqrt{a}+2\right)}\)

Ta có: \(A=\left(\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a+\sqrt{a}}{a-1}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{1}{\sqrt{a}-1}\right)\)

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

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

\(=\dfrac{\sqrt{a}+1}{2\sqrt{a}}\)

26 tháng 12 2021

a: \(A=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-1-a+4}\)

\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)

27 tháng 12 2021

\(ĐK:a>0;a\ne1;a\ne4\\ a,A=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\\ b,A>0\Leftrightarrow\sqrt{a}-2>0\Leftrightarrow a>4\)

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

8 tháng 1 2022

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

17 tháng 7 2021

Làm ơn giúp mình với... :(

a: Ta có: \(A=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right)\cdot\dfrac{x-4}{3\sqrt{x}}\)

\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{x-4}{3\sqrt{x}}\)

\(=\dfrac{2}{3}\)

 

\(A=\dfrac{2-\sqrt{a}-\sqrt{a}-3}{2\sqrt{a}+1}=-1\)

\(B=\dfrac{1}{1-\sqrt{2+\sqrt{3}}}-\dfrac{1}{1-\sqrt{2-\sqrt{3}}}\)

\(=\dfrac{\sqrt{2}}{\sqrt{2}-\sqrt{3}-1}-\dfrac{\sqrt{2}}{\sqrt{2}-\sqrt{3}+1}\)

\(=\dfrac{2-\sqrt{6}+\sqrt{2}-2+\sqrt{6}+\sqrt{2}}{5-2\sqrt{6}-1}\)

\(=\dfrac{2\sqrt{2}}{4-2\sqrt{6}}=\dfrac{1}{\sqrt{2}-\sqrt{3}}=-\sqrt{2}-\sqrt{3}\)