Rút gọn nhá!!!!!!
\(\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}=????????\)
K
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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: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
30 tháng 9 2016
\(=\frac{-2\sqrt{a}}{\left(a-1\right)\left(\sqrt{a}+1\right)}.\frac{\left(1-a\right)^2}{2}=\frac{-2\sqrt{a}}{-\left(1-a\right)\left(\sqrt{a}+1\right)}.\frac{\left(1-a\right)^2}{2}=\frac{\sqrt{a}\left(1-a\right)}{\sqrt{a}+1}.\)
\(=\frac{\sqrt{a}\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}{\sqrt{a}+1}=\sqrt{a}\left(1-\sqrt{a}\right)\)
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
\(A=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right).\)
\(A=\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1-a+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(A=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)\)
\(A=\frac{\sqrt{a}-2}{\sqrt{a}}\)
9 tháng 7 2019
\(A=\)\(\frac{\sqrt{a}\left(1-a\right)^2}{1+a}:\left[\left(\frac{1-\sqrt{a}^3}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1+\sqrt{a}^3}{1+\sqrt{a}}-\sqrt{a}\right)\right]\)
\(=\frac{\sqrt{a}\left(1-a\right)^2}{1+a}\)\(:\)\(\left[\left(1+\sqrt{a}+a+\sqrt{a}\right)\left(1-\sqrt{a}+a-\sqrt{a}\right)\right]\)
\(=\frac{\sqrt{a}\left(1-a\right)^2}{1+a}:\)\(\left(1+a+2\sqrt{a}\right)\left(1+a-2\sqrt{a}\right)\)
\(=\frac{\sqrt{a}\left(1-a\right)^2}{\left(1+a\right)\left[\left(1+a\right)^2-\left(2\sqrt{a}\right)^2\right]}\)\(=\frac{\sqrt{a}\left(1-a\right)^2}{\left(a+1\right)\left(1+2a+a^2-4a\right)}\)
\(=\frac{\sqrt{a}\left(1-a\right)^2}{\left(a+1\right)\left(1-a\right)^2}=\frac{\sqrt{q}}{a+1}\)
\(\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}=\sqrt{\left(1+\frac{1}{a}-\frac{1}{a+1}\right)^2}=\left|1+\frac{1}{a}-\frac{1}{a+1}\right|\)
Đây xem nhá
\(\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}=\sqrt{\frac{\left(a\left(a+1\right)\right)^2+\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}}=\sqrt{\frac{\left(a\left(a+1\right)\right)^2+a^2+2a+1+a^2}{a^2\left(a+1\right)^2}}=\sqrt{\frac{\left(a\left(a+1\right)\right)^2+2a\left(a+1\right)+1^2}{a^2\left(a+1\right)^2}}\)
\(=\sqrt{\left(\frac{a\left(a+1\right)+1}{a\left(a+1\right)}\right)^2}=\left|\frac{a\left(a+1\right)+1}{a\left(a+1\right)}\right|\)