Rút gọn biểu thức sau \(P=\left(\frac{\sqrt{a}+2}{a+2\sqrt{a}+1}-\frac{\sqrt{a}-2}{a-1}\right).\frac{\sqrt{a}+1}{\sqrt{a}}\)
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10 tháng 8 2017
\(A=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\frac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\sqrt{a}}:\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{1}{\left(\sqrt{a}-1\right)\sqrt{a}}:\frac{a-1-\left(a-4\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{1}{\left(\sqrt{a}-1\right)\sqrt{a}}:\frac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{1}{\left(\sqrt{a}-1\right)\sqrt{a}}.\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
BT
10 tháng 8 2017
\(A=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right).\)
\(A=\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1-a+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(A=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)\)
\(A=\frac{\sqrt{a}-2}{\sqrt{a}}\)
12 tháng 12 2016
Điều kiện: \(\hept{\begin{cases}a>0\\\sqrt{a}-1\ne0\\\sqrt{a}-2\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}a>0\\a\ne1\\a\ne4\end{cases}}\)
Ta có:
\(1P=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1-a+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)\)
\(=\frac{\sqrt{a}-2}{\sqrt{a}}\)
KT
11 tháng 7 2018
Bài 1:
a) \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b) \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)
\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)
c) ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
\(ĐKXĐ:a>1\)
\(P=\left(\frac{\sqrt{a}+2}{a+2\sqrt{a}+1}-\frac{\sqrt{a}-2}{a-1}\right)\cdot\frac{\sqrt{a}+1}{\sqrt{a}}\)
\(\Leftrightarrow P=\left(\frac{\sqrt{a}+2}{\left(\sqrt{a}+1\right)^2}-\frac{\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\cdot\frac{\sqrt{a}+1}{\sqrt{a}}\)
\(\Leftrightarrow P=\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}\cdot\frac{\sqrt{a}+1}{\sqrt{a}}\)
\(\Leftrightarrow P=\frac{a+\sqrt{a}-2-a+\sqrt{a}+2}{\sqrt{a}\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(\Leftrightarrow P=\frac{2\sqrt{a}}{\sqrt{a}\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(\Leftrightarrow P=\frac{2}{a-1}\)
\(ĐKXĐ:\hept{\begin{cases}a>0\\a\ne1\end{cases}}\)
Ta có :
\(P=\left(\frac{\sqrt{a}+2}{a+2\sqrt{a}+1}-\frac{\sqrt{a}-2}{a-1}\right).\frac{\sqrt{a}+1}{\sqrt{a}}\)
\(=\left(\frac{\sqrt{a}+2}{\left(\sqrt{a}+1\right)^2}-\frac{\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right).\frac{\sqrt{a}+1}{\sqrt{a}}\)
\(=\left(\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}\right).\frac{\sqrt{a}+1}{\sqrt{a}}\)
\(=\left(\frac{\left(a+\sqrt{a}-2\right)-\left(a-\sqrt{a}-2\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}\right).\frac{\sqrt{a}+1}{\sqrt{a}}\)
\(=\frac{2\sqrt{a}}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}.\frac{\sqrt{a}+1}{\sqrt{a}}\)
\(=\frac{2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{2}{a-1}\)
Vậy \(P=\frac{2}{a-1}\left(a>0;a\ne1\right)\)