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\(1+\left(\frac{a+2\sqrt{a}-1}{1-a}-\frac{2a\sqrt{a}-\sqrt{a}+a}{1-a\sqrt{a}}\right)\cdot\frac{a-\sqrt{a}}{2\sqrt{a}-1}\)
\(=1+\left(\frac{\left(\sqrt{a}-1\right)^2}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}-\frac{\sqrt{a}\left(1+\sqrt{a}+a\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}\right)\cdot\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)
\(=1+\left(\frac{\left(1-\sqrt{a}\right)^2}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}-\frac{\sqrt{a}}{\left(1-\sqrt{a}\right)}\right)\cdot\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)
\(=1+\left(\frac{\left(1-\sqrt{a}\right)}{\left(1+\sqrt{a}\right)}-\frac{\sqrt{a}}{\left(1-\sqrt{a}\right)}\right)\cdot\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)
\(=1+\left(\frac{\left(1-\sqrt{a}\right)^2}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}-\frac{\sqrt{a}\left(1+\sqrt{a}\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right)\cdot\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)
\(=1+\frac{1-2\sqrt{a}+a-\sqrt{a}-a}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\cdot\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)
\(=1+\frac{1-2\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\cdot\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)
\(=1+\frac{1-2\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\cdot\frac{\sqrt{a}\left(1-\sqrt{a}\right)}{1-2\sqrt{a}}\)
\(=1+\frac{\sqrt{a}}{\left(1+\sqrt{a}\right)}\)
\(=\frac{1+\sqrt{a}+\sqrt{a}}{1+\sqrt{a}}\)
\(=\frac{1+2\sqrt{a}}{1+\sqrt{a}}\)
ĐKXĐ: \(x\ge0\)
a/ \(Q=1+\left[\frac{\left(2\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)}-\frac{2a\sqrt{a}-\sqrt{a}+a}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}\right].\frac{\sqrt{a}.\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)
\(=1+\left[\frac{2\sqrt{a}-1}{1-\sqrt{a}}-\frac{2a\sqrt{a}-\sqrt{a}+a}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}\right].\frac{\sqrt{a}.\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)
\(=1+\left[\frac{\left(2\sqrt{a}-1\right)\left(1+\sqrt{a}+a\right)-\left(2a\sqrt{a}-\sqrt{a}+a\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}\right].\frac{\sqrt{a}.\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)
\(=1+\left[\frac{2\sqrt{a}-1}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}\right].\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)\(=1+\frac{\sqrt{a}}{1+\sqrt{a}+a}=\frac{1+2\sqrt{a}+a}{1+\sqrt{a}+a}\)
b/ \(Q=\frac{\sqrt{6}}{1+\sqrt{6}}\Leftrightarrow1+\frac{\sqrt{a}}{1+\sqrt{a}+a}=\frac{\sqrt{6}}{1+\sqrt{6}}\) \(\Leftrightarrow\frac{1+2\sqrt{a}+a}{1+\sqrt{a}+a}=\frac{\sqrt{6}}{1+\sqrt{6}}\)
\(\Leftrightarrow\left(1+\sqrt{6}\right)\left(1+2\sqrt{a}+a\right)=\sqrt{6}\left(1+\sqrt{a}+a\right)\)
Tới đây mình k biết giải
c/ \(Q>\frac{2}{3}\Leftrightarrow\frac{1+2\sqrt{a}+a}{1+\sqrt{a}+a}>\frac{2}{3}\)\(\Leftrightarrow3.\left(1+2\sqrt{a}+a\right)>2.\left(1+\sqrt{a}+a\right)\)
\(\Leftrightarrow3+6\sqrt{a}+3a-2-2\sqrt{a}-2a>0\Leftrightarrow1+4\sqrt{a}+a>0\)
\(\Leftrightarrow x< -2-\sqrt{3}Vx>-2+\sqrt{3}\)