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

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

\(=\dfrac{4\sqrt{a}\left(1+a-1\right)}{\sqrt{a}}\)

\(=4a\)

Để \(\sqrt{A}>A\) thì \(\sqrt{4a}>4a\)

\(\Leftrightarrow2\sqrt{a}-4a>0\)

\(\Leftrightarrow2\sqrt{a}\left(1-2\sqrt{a}\right)>0\)

\(\Leftrightarrow2\sqrt{a}< 1\)

\(\Leftrightarrow a< \dfrac{1}{4}\)

Kết hợp ĐKXĐ, ta được: \(0< a< \dfrac{1}{4}\)

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

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

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

24 tháng 6 2017

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

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

\(=\left(\dfrac{2\cdot2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+4\sqrt{a}\right)\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}\)

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

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

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

\(=4\left(1+a-1\right)\)

\(=4a\)

Để \(\sqrt{a}>A\) thì \(\sqrt{a}>4a\)

\(\Leftrightarrow a>\sqrt{4a}\left(đk:a\ge0\right)\)

\(\Leftrightarrow a>2\sqrt{a}\)

\(\Leftrightarrow2\sqrt{a}< a\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{a}< a\left(đk:a\ge0\right)\\2\sqrt{a}< a\left(đk:a< 0\right)\end{matrix}\right.\)

23 tháng 8 2017

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

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

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

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

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

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

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

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

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

9 tháng 6 2017

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

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

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

biểu thức trong dấu ngoặc vuông = \(\left[\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)\right]^2=\left(a-1\right)^2\)

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

16 tháng 11 2021

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