\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)....\)
a) Rút gọn A, b) tìm a để A<0.c) Tìm A để A=-2
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13 tháng 8 2019
\(đkxđ\Leftrightarrow x\ge0;x\ne1;x\ne4\)
\(A=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right).\)
\(=\left(\frac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\)\(\left(\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\left(\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3\sqrt{a}\left(\sqrt{a}-1\right)}=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
\(A< \frac{1}{6}\Rightarrow\frac{\sqrt{a}-2}{3\sqrt{a}}>\frac{1}{6}\Rightarrow\frac{\sqrt{a}-2}{3\sqrt{a}}-\frac{1}{6}>0\)
\(\Rightarrow\frac{2\left(\sqrt{a}-2\right)}{6\sqrt{a}}-\frac{\sqrt{a}}{6\sqrt{a}}>0\Rightarrow\frac{\sqrt{a}-4}{6\sqrt{a}}>0\)
Vì \(6\sqrt{a}>0\Rightarrow\sqrt{a}-4>0\Rightarrow\sqrt{a}>4\Rightarrow a>16\)
Vậy \(P>\frac{1}{6}\Leftrightarrow a>16\)
\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\).\(\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
= \(\left[\left(\frac{\sqrt{a}}{2}\right)^2-2\frac{\sqrt{a}}{2}\frac{1}{2\sqrt{a}}+\left(\frac{1}{2\sqrt{a}}\right)^2\right]\).\(\left[\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-1\right)}{a-1}\cdot\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+1\right)}{a-1}\right]\)
=\(\left(\frac{a}{4}-\frac{1}{2}+\frac{1}{4a}\right)\).\(\left[\frac{\left(\sqrt{a}-1\right)^2}{a-1}\cdot\frac{\left(\sqrt{a}+1\right)^2}{a-1}\right]\)
=\(\left(\frac{a^2}{4a}-\frac{2a}{4a}+\frac{1}{4a}\right)\).\(\left[\frac{\left[\left(\sqrt{a}-1\right)-\left(\sqrt{a}+1\right)\right]\cdot\left[\left(\sqrt{a}-1\right)+\left(\sqrt{a}+1\right)\right]}{a-1}\right]\)
=\(\left(\frac{a^2-2a+1}{4a}\right)\).\(\left[\frac{\left(\sqrt{a}-1-\sqrt{a}+1\right).\left(\sqrt{a}-1+\sqrt{a}+1\right)}{a-1}\right]\)
=\(\frac{\left(a-1\right)^2}{1}\).\(\frac{-4\sqrt{a}}{a-1}\)
=\(\frac{-\left(a-1\right)}{1}\)= - a + 1
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