\(\frac{1+\sqrt{1-a}}{1-a+\sqrt{1-a}}+\frac{1-\sqrt{1+a}}{1+a-\sqrt{1+a}}+\frac{1}{\sqrt{1+a}}\)...">
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11 tháng 8 2020

A= \(\left(\frac{\sqrt{b}}{a-\sqrt{ab}}-\frac{\sqrt{a}}{\sqrt{ab}-b}\right).\left(a\sqrt{b}-b\sqrt{a}\right)\)

A = \(\left(\frac{\sqrt{b}}{\sqrt{a}.\sqrt{a}-\sqrt{ab}}-\frac{\sqrt{a}}{\sqrt{ab}-\sqrt{b}.\sqrt{b}}\right).\left(a\sqrt{b}-b\sqrt{a}\right)\)

A = \(\left(\frac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}\right).\left(a\sqrt{b}-b\sqrt{a}\right)\)

A = \(\left(\frac{b}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\right)\left(a\sqrt{b}-b\sqrt{a}\right)\)

A = \(\left(\frac{b-a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\right).\left(\sqrt{a}.\sqrt{a}.\sqrt{b}-\sqrt{b}.\sqrt{b}\sqrt{a}\right)\)

A = \(\left(\frac{b-a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\right).\left(\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\right)\)

A = b-a

B = \(\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{\sqrt{a}}{a-\sqrt{a}}\right):\frac{\sqrt{a}+1}{a-1}\)

B = \(\left(\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{a-1}-\frac{\sqrt{a}\left(a+\sqrt{a}\right)}{a^2-a}\right).\frac{a-1}{\sqrt{a}+1}\)

B = \(\left(\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{a-1}-\frac{\sqrt{a}.\sqrt{a}\left(\sqrt{a}+1\right)}{a\left(a-1\right)}\right).\frac{a-1}{\sqrt{a}+1}\)

\(B=\left(\frac{a\sqrt{a}\left(\sqrt{a}+1\right)}{a\left(a-1\right)}-\frac{a\left(\sqrt{a}+1\right)}{a\left(a-1\right)}\right).\frac{a-1}{\sqrt{a}+1}\)

B= \(\left(\frac{a\sqrt{a}\left(\sqrt{a}+1\right)-a\left(\sqrt{a}+1\right)}{a\left(a-1\right)}\right).\frac{a-1}{\sqrt{a}+1}\)

B= \(\left(\frac{\left(\sqrt{a}+1\right)\left(a\sqrt{a}-a\right)}{a\left(a-1\right)}\right).\frac{a-1}{\sqrt{a}+1}\)

B = \(\frac{\left(\sqrt{a}+1\right)a\left(\sqrt{a}-1\right)}{a\left(a-1\right)}.\frac{a-1}{\sqrt{a}+1}\)

\(B=\frac{a\left(\sqrt{a}^2-1^2\right)}{a\left(a-1\right)}.\frac{a-1}{\sqrt{a}+1}\)

\(B=\frac{a\left(a-1\right)}{a\left(a-1\right)}.\frac{a-1}{\sqrt{a}+1}\)

B = \(\frac{a-1}{\sqrt{a}+1}\)

25 tháng 4 2020

thanks

7 tháng 9 2020

a) \(\frac{2\sqrt{5}+b\sqrt{a}}{\sqrt{ab}}-\frac{1}{\sqrt{a}-\sqrt{b}}\)

\(=\frac{2\sqrt{5ab}+ab\sqrt{b}}{ab}-\frac{\sqrt{a}+\sqrt{b}}{a-b}\)

\(=\frac{\left(2\sqrt{5ab}+ab\sqrt{b}\right)\left(a-b\right)-\left(\sqrt{a}+\sqrt{b}\right)ab}{ab\left(a-b\right)}\)

\(=\frac{2a\sqrt{5ab}-2b\sqrt{5ab}+a^2b\sqrt{b}-ab^2\sqrt{b}-ab\sqrt{a}-ab\sqrt{b}}{ab\left(a-b\right)}\)

7 tháng 9 2020

b) \(\left(1+\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\cdot\left(\frac{1+1+\sqrt{a}}{1-a^2}\right)^2\)

\(=\left(1+\frac{\left(\sqrt{a}-1\right)\sqrt{a}}{\sqrt{a}-1}\right)\cdot\left(\frac{2+\sqrt{a}}{\left(1-a\right)\left(1+a\right)}\right)^2\)

\(=\left(1+\sqrt{a}\right)\left[\frac{2+\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\left(1+a\right)}\right]^2\)

\(=\frac{4+4\sqrt{a}+a}{\left(1-\sqrt{a}\right)^2\left(1+\sqrt{a}\right)\left(1+a\right)^2}\)  \(a\ge0;a\ne\pm1\)

Bổ sung đk của phần a: \(a,b>0;a\ne b\)

18 tháng 6 2019

\(A=\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}\\ A=\frac{\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{2\left(1+\sqrt{2}\right)}{1+\sqrt{2}}\\ A=\sqrt{3}-2\)

\(C=\left(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{a+2}{a-2}\\ C=\left(\frac{a\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{a\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right):\frac{a+2}{a-2}\\ C=\left(\frac{a}{\sqrt{a}}-\frac{a}{\sqrt{a}}\right):\frac{a+2}{a-2}\\ C=0\)

18 tháng 6 2019

Mình thấy hình như chưa đúng lắm

8 tháng 11 2021

\(A=\left(a-1\right)\sqrt{\frac{a}{a-1}}+\sqrt{a\left(a-1\right)}-a\sqrt{\frac{a-1}{a}}\)

\(A=\sqrt{\left(a-1\right)^2.\frac{a}{a-1}}+\sqrt{a\left(a-1\right)}-\sqrt{a^2.\frac{a-1}{a}}\)

\(A=\sqrt{\left(a-1\right)a}+\sqrt{a\left(a-1\right)}-\sqrt{a\left(a-1\right)}\)

\(A=\sqrt{a\left(a-1\right)}\)

5 tháng 8 2019

chỗ có 2 số 1 bỏ hộ mk 1 số nha

5 tháng 8 2019

Tui cx đang có câu như thế mà k bt làm đây

Hu hu

1 tháng 11 2020

\(M=\left(\frac{2+\sqrt{a}}{\left(\sqrt{a}+1\right)^2}-\frac{\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\frac{a\left(\sqrt{a}+1\right)-\left(\sqrt{a}+1\right)}{a}\)

\(=\frac{\left(2+\sqrt{a}\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(a-1\right)}{a}\)

\(=\frac{2\sqrt{a}-2+a-\sqrt{a}-a-\sqrt{a}+2\sqrt{a}+2}{\left(\sqrt{a}+1\right)\left(a-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(a-1\right)}{a}\)

\(=\frac{2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(a-1\right)}{a}\)

\(=\frac{2\sqrt{a}\left(\sqrt{a-1}\right)}{a\left(\sqrt{a}+1\right)}=\frac{2\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\)

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