rút gọn bt sau:
\(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a-1}}:\frac{\sqrt{a+1}}{a-2\sqrt{a+1}}\)
K
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TH
1 tháng 11 2020
\(M=\left(\frac{2+\sqrt{a}}{\left(\sqrt{a}+1\right)^2}-\frac{\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\frac{a\left(\sqrt{a}+1\right)-\left(\sqrt{a}+1\right)}{a}\)
\(=\frac{\left(2+\sqrt{a}\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(a-1\right)}{a}\)
\(=\frac{2\sqrt{a}-2+a-\sqrt{a}-a-\sqrt{a}+2\sqrt{a}+2}{\left(\sqrt{a}+1\right)\left(a-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(a-1\right)}{a}\)
\(=\frac{2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(a-1\right)}{a}\)
\(=\frac{2\sqrt{a}\left(\sqrt{a-1}\right)}{a\left(\sqrt{a}+1\right)}=\frac{2\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\)
TH
1 tháng 11 2020
\(N=\left(\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2}{a-1}+4\sqrt{a}\right)\cdot\frac{a-1}{\sqrt{a}}\)
\(=\left(\frac{a+1+2\sqrt{a}-a-1+2\sqrt{a}}{a-1}+4\sqrt{a}\right)\cdot\frac{a-1}{\sqrt{a}}\)
\(=\left(\frac{4\sqrt{a}}{a-1}+4\sqrt{a}\right)\cdot\frac{a-1}{\sqrt{a}}=4\sqrt{a}\left(\frac{1}{a-1}+1\right)\cdot\frac{a-1}{\sqrt{a}}=4\cdot\left(a-1\right)\left(\frac{1}{a-1}+1\right)\)
\(=4\cdot\left(a-1\right)\)
vừa tham khảo cách làm vừa check lại hộ tớ với nhé :33
NM
2
21 tháng 9 2015
ĐK a < 0 hoặc a = 1
\(=\sqrt{1-a}+\sqrt{a\left(a-1\right)}-\sqrt{\frac{a^2\left(a-1\right)}{a}}=\sqrt{1-a}+\sqrt{a\left(a-1\right)}-\sqrt{a\left(a-1\right)}=\sqrt{1-a}\)
21 tháng 9 2015
Đk \(1-a\ge0;a\left(1-a\right)\ge0;\frac{a}{a-1}\ge0\)
=> \(a\le1;a\ge1ora\le0;a\ge1ora
MT
6 tháng 9 2015
\(B=\left(\frac{1}{1-\sqrt{a}}-\frac{1}{1+\sqrt{a}}\right)\left(\frac{1}{\sqrt{a}}+1\right)\)
\(=\left(\frac{1+\sqrt{a}}{1-a}-\frac{1-\sqrt{a}}{1-a}\right)\left(\frac{\sqrt{a}}{a}+\frac{a}{a}\right)\)
\(=\frac{1+\sqrt{a}-1+\sqrt{a}}{1-a}.\frac{\sqrt{a}+a}{a}\)
\(=\frac{2\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}.\frac{\sqrt{a}.\left(1+\sqrt{a}\right)}{a}\)
\(=\frac{2}{1-\sqrt{a}}\)
NY
0
MN
29 tháng 3 2020
\(ĐKXĐ:a\ge0\)
\(A=\left(\frac{2\sqrt{a}}{a\sqrt{a}+a+\sqrt{a}+1}+\frac{1}{\sqrt{a}+1}\right):\left(1+\frac{\sqrt{a}}{a+1}\right)\)
\(\Leftrightarrow A=\left(\frac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}+1\right)}+\frac{1}{\sqrt{a}+1}\right):\frac{a+\sqrt{a}+1}{a+1}\)
\(\Leftrightarrow A=\frac{2\sqrt{a}+a+1}{\left(a+1\right)\left(\sqrt{a}+1\right)}\cdot\frac{a+1}{a+\sqrt{a}+1}\)
\(\Leftrightarrow A=\frac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(a+\sqrt{a}+1\right)}\)
\(\Leftrightarrow A=\frac{\sqrt{a}+1}{a+\sqrt{a}+1}\)