Rút gọn \(A=\sqrt{1-a}+\sqrt{a\left(a-1\right)}+a\sqrt{\frac{a-1}{a}}\)
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9 tháng 7 2019
\(A=\)\(\frac{\sqrt{a}\left(1-a\right)^2}{1+a}:\left[\left(\frac{1-\sqrt{a}^3}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1+\sqrt{a}^3}{1+\sqrt{a}}-\sqrt{a}\right)\right]\)
\(=\frac{\sqrt{a}\left(1-a\right)^2}{1+a}\)\(:\)\(\left[\left(1+\sqrt{a}+a+\sqrt{a}\right)\left(1-\sqrt{a}+a-\sqrt{a}\right)\right]\)
\(=\frac{\sqrt{a}\left(1-a\right)^2}{1+a}:\)\(\left(1+a+2\sqrt{a}\right)\left(1+a-2\sqrt{a}\right)\)
\(=\frac{\sqrt{a}\left(1-a\right)^2}{\left(1+a\right)\left[\left(1+a\right)^2-\left(2\sqrt{a}\right)^2\right]}\)\(=\frac{\sqrt{a}\left(1-a\right)^2}{\left(a+1\right)\left(1+2a+a^2-4a\right)}\)
\(=\frac{\sqrt{a}\left(1-a\right)^2}{\left(a+1\right)\left(1-a\right)^2}=\frac{\sqrt{q}}{a+1}\)
T
2 tháng 8 2017
Điều kiện : a> 0 ; a khác 1
\(A=\frac{\left(\sqrt{a}\right)^3-1}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}\right)^3+1}{\sqrt{a}\left(\sqrt{a}+1\right)}+\left(\frac{a-1}{\sqrt{a}}\right)\left(\frac{\left(\sqrt{a}+1\right)^2+\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(A=\frac{a+\sqrt{a}+1}{\sqrt{a}}-\frac{a-\sqrt{a}+1}{\sqrt{a}}+\left(\frac{a-1}{\sqrt{a}}\right)\left(\frac{2a+2}{a-1}\right)\)
\(A=\frac{2\sqrt{a}}{\sqrt{a}}+\frac{2\left(a+1\right)}{\sqrt{a}}=2+\frac{2\sqrt{a}\left(a+1\right)}{a}\)
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
\(A=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right).\)
\(A=\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1-a+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(A=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)\)
\(A=\frac{\sqrt{a}-2}{\sqrt{a}}\)