Cho biểu thức: \(P=1+\left(\frac{2a+\sqrt{a}-1}{1-a}-\frac{2a\sqrt{a}-\sqrt{a}+a}{1-a\sqrt{a}}\right).\frac{a-\sqrt{a}}{2\sqrt{a}-1}\)
a) Rút gọn P
b) Cho \(P=\frac{\sqrt{6}}{1+\sqrt{6}}\). Tìm giá trị của a. Chứng minh rằng P > \(\frac{2}{3}\)
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BA
2 tháng 8 2019
\(1+\left(\frac{a+2\sqrt{a}-1}{1-a}-\frac{2a\sqrt{a}-\sqrt{a}+a}{1-a\sqrt{a}}\right)\cdot\frac{a-\sqrt{a}}{2\sqrt{a}-1}\)
\(=1+\left(\frac{\left(\sqrt{a}-1\right)^2}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}-\frac{\sqrt{a}\left(1+\sqrt{a}+a\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}\right)\cdot\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)
\(=1+\left(\frac{\left(1-\sqrt{a}\right)^2}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}-\frac{\sqrt{a}}{\left(1-\sqrt{a}\right)}\right)\cdot\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)
\(=1+\left(\frac{\left(1-\sqrt{a}\right)}{\left(1+\sqrt{a}\right)}-\frac{\sqrt{a}}{\left(1-\sqrt{a}\right)}\right)\cdot\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)
\(=1+\left(\frac{\left(1-\sqrt{a}\right)^2}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}-\frac{\sqrt{a}\left(1+\sqrt{a}\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right)\cdot\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)
\(=1+\frac{1-2\sqrt{a}+a-\sqrt{a}-a}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\cdot\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)
\(=1+\frac{1-2\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\cdot\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)
\(=1+\frac{1-2\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\cdot\frac{\sqrt{a}\left(1-\sqrt{a}\right)}{1-2\sqrt{a}}\)
\(=1+\frac{\sqrt{a}}{\left(1+\sqrt{a}\right)}\)
\(=\frac{1+\sqrt{a}+\sqrt{a}}{1+\sqrt{a}}\)
\(=\frac{1+2\sqrt{a}}{1+\sqrt{a}}\)
20 tháng 8 2017
\(A=1+"\frac{2a+\sqrt{a}-1}{1-a}-\frac{2a\sqrt{a}-\sqrt{a}+a}{1-a\sqrt{a}}"\times\frac{a-\sqrt{a}}{2\sqrt{a}-1}=\)
\(A="\frac{1a+\sqrt{a}-1}{1-a}-\frac{1a\sqrt{a}-\sqrt{a}+a}{1-a\sqrt{a}}"\times\frac{a-\sqrt{a}}{1\sqrt{a}-1}\)
P/s: Ko chắc đâu nhé
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 \(P>\frac{1}{6}\Leftrightarrow a>16\)
10 tháng 8 2015
Điều kiện: x \(\ne\) 1; 1/4 ; x \(\ge\) 0
\(A=1+\left(\frac{\left(2a+\sqrt{a}-1\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}-\frac{\left(2a+\sqrt{a}-1\right).\sqrt{a}}{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\right)\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\right)\)
\(A=1+\left(\frac{\left(2a+\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)-\left(2a+\sqrt{a}-1\right)\left(1+\sqrt{a}\right).\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\right)\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\right)\)
\(A=1+\left(\frac{\left(2a+\sqrt{a}-1\right)\left(a+\sqrt{a}+1-a-\sqrt{a}\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\right)\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\right)\)
\(A=1+\left(\frac{\left(2a+\sqrt{a}-1\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\right)\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\right)\)
\(A=1+\left(\frac{\left(2\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\right)\left(\frac{-\sqrt{a}\left(1-\sqrt{a}\right)}{2\sqrt{a}-1}\right)=1+\frac{-\sqrt{a}}{a+\sqrt{a}+1}=\frac{a+1}{a+\sqrt{a}+1}\)
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