Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+1}+\frac{2}{a-1}\right)\left(a>0;a\ne1\right)\)
\(A=\frac{\sqrt{a}.\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{\left(\sqrt{a}-1\right)+2}{a-1}\)
\(A=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{\sqrt{a}+1}{a-1}\)
\(A=\frac{\sqrt{a}+1}{\sqrt{a}}:\frac{1}{\sqrt{a}-1}\)
\(A=\frac{\sqrt{a}+1}{\sqrt{a}}.\left(\sqrt{a}-1\right)=\frac{a-1}{\sqrt{a}}\)
Vậy..............
\(B=\left(\frac{\sqrt{a}}{\sqrt{a}+1}-\frac{\sqrt{a}}{\sqrt{a}-1}+\frac{1}{a-1}\right):\frac{a}{2+2\sqrt{a}}\)( điều kiện như trên )
\(B=\frac{\sqrt{a}\left(\sqrt{a}-1\right)-\sqrt{a}\left(\sqrt{a}+1\right)+1}{a-1}:\frac{a}{2\left(1+\sqrt{a}\right)}\)
\(B=\frac{a-\sqrt{a}-a-\sqrt{a}+1}{a-1}:\frac{a}{\left(\sqrt{a}+1\right).2}\)
\(B=\frac{1-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}.\frac{\left(\sqrt{a}+1\right).2}{a}\)
\(B=\frac{2\left(1-2\sqrt{a}\right)}{a\left(\sqrt{a}-1\right)}\)
Vậy.........
_Minh ngụy_
\(A=\)\(\frac{\sqrt{a}\left(1-a\right)^2}{1+a}:\left[\left(\frac{1-\sqrt{a}^3}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1+\sqrt{a}^3}{1+\sqrt{a}}-\sqrt{a}\right)\right]\)
\(=\frac{\sqrt{a}\left(1-a\right)^2}{1+a}\)\(:\)\(\left[\left(1+\sqrt{a}+a+\sqrt{a}\right)\left(1-\sqrt{a}+a-\sqrt{a}\right)\right]\)
\(=\frac{\sqrt{a}\left(1-a\right)^2}{1+a}:\)\(\left(1+a+2\sqrt{a}\right)\left(1+a-2\sqrt{a}\right)\)
\(=\frac{\sqrt{a}\left(1-a\right)^2}{\left(1+a\right)\left[\left(1+a\right)^2-\left(2\sqrt{a}\right)^2\right]}\)\(=\frac{\sqrt{a}\left(1-a\right)^2}{\left(a+1\right)\left(1+2a+a^2-4a\right)}\)
\(=\frac{\sqrt{a}\left(1-a\right)^2}{\left(a+1\right)\left(1-a\right)^2}=\frac{\sqrt{q}}{a+1}\)
\(M=\left(\frac{2+\sqrt{a}}{\left(\sqrt{a}+1\right)^2}-\frac{\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\frac{a\left(\sqrt{a}+1\right)-\left(\sqrt{a}+1\right)}{a}\)
\(=\frac{\left(2+\sqrt{a}\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(a-1\right)}{a}\)
\(=\frac{2\sqrt{a}-2+a-\sqrt{a}-a-\sqrt{a}+2\sqrt{a}+2}{\left(\sqrt{a}+1\right)\left(a-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(a-1\right)}{a}\)
\(=\frac{2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(a-1\right)}{a}\)
\(=\frac{2\sqrt{a}\left(\sqrt{a-1}\right)}{a\left(\sqrt{a}+1\right)}=\frac{2\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\)
\(N=\left(\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2}{a-1}+4\sqrt{a}\right)\cdot\frac{a-1}{\sqrt{a}}\)
\(=\left(\frac{a+1+2\sqrt{a}-a-1+2\sqrt{a}}{a-1}+4\sqrt{a}\right)\cdot\frac{a-1}{\sqrt{a}}\)
\(=\left(\frac{4\sqrt{a}}{a-1}+4\sqrt{a}\right)\cdot\frac{a-1}{\sqrt{a}}=4\sqrt{a}\left(\frac{1}{a-1}+1\right)\cdot\frac{a-1}{\sqrt{a}}=4\cdot\left(a-1\right)\left(\frac{1}{a-1}+1\right)\)
\(=4\cdot\left(a-1\right)\)
vừa tham khảo cách làm vừa check lại hộ tớ với nhé :33