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Ariels spring fashion
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Tô Hoài An
1 tháng 11 2020 lúc 20:21

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

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

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

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

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

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Tô Hoài An
1 tháng 11 2020 lúc 20:29

\(N=\left(\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2}{a-1}+4\sqrt{a}\right)\cdot\frac{a-1}{\sqrt{a}}\)

\(=\left(\frac{a+1+2\sqrt{a}-a-1+2\sqrt{a}}{a-1}+4\sqrt{a}\right)\cdot\frac{a-1}{\sqrt{a}}\)

\(=\left(\frac{4\sqrt{a}}{a-1}+4\sqrt{a}\right)\cdot\frac{a-1}{\sqrt{a}}=4\sqrt{a}\left(\frac{1}{a-1}+1\right)\cdot\frac{a-1}{\sqrt{a}}=4\cdot\left(a-1\right)\left(\frac{1}{a-1}+1\right)\)

\(=4\cdot\left(a-1\right)\)

vừa tham khảo cách làm vừa check lại hộ tớ với nhé :33 

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Nguyễn Khánh Ly
1 tháng 11 2020 lúc 20:35
\(Với\)\(a>0\);\(a\ne1\)ta có:

\(M=(\frac{2+\sqrt{a}}{a+2\sqrt{a}+1}-\frac{\sqrt{a}-2}{a-1}).(\frac{a\sqrt{a}+a-\sqrt{a}-1}{\sqrt{a}})\)

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

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

\(=[\frac{a+\sqrt{a}-2}{(\sqrt{a}+1)(a-1)}-\frac{a-\sqrt{a}-2}{(\sqrt{a}+1)(a-1)}].\frac{(a-1).(\sqrt{a}+1)}{\sqrt{a}}\)

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

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

\(=2\)

Vậy \(M=2\)

\(Với\)\(a>0;a\ne1:\)

\(N=(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}).(\sqrt{a}-\frac{1}{\sqrt{a}})\)

\(=[\frac{(\sqrt{a}+1).(\sqrt{a}+1)}{\left(\sqrt{a}-1\right).(\sqrt{a}+1)}-\frac{(\sqrt{a}-1).(\sqrt{a}-1)}{(\sqrt{a}-1).(\sqrt{a}+1)}+\frac{4\sqrt{a}(a-1)}{(\sqrt{a}-1).(\sqrt{a}+1)}].\frac{a-1}{\sqrt{a}}\)

\(=\frac{(\sqrt{a}+1)^2-(\sqrt{a}-1)^2+(4a\sqrt{a}-4\sqrt{a})}{(\sqrt{a}-1).(\sqrt{a}+1)}.\frac{a-1}{\sqrt{a}}\)

\(=\frac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4a\sqrt{a}-4\sqrt{a}}{a-1}.\frac{a-1}{\sqrt{a}}\)

\(=\frac{4a\sqrt{a}}{a-1}.\frac{a-1}{\sqrt{a}}\)\(=4a\)

Vậy \(N=4a\)

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Như Ý
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nguyen le duy hung
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Không Tên
11 tháng 7 2018 lúc 20:04

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:  \(a\ge0;a\ne1\)

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

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

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

\(=1-a+a=1\)

Dragon Boy
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\(A=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+1}+\frac{2}{a-1}\right)\left(a>0;a\ne1\right)\)

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

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

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

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

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_

Dragon Boy
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Trần Nguyễn
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Nguyen hoang chi
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Nguyễn Nhã Thanh
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Đinh Đức Hùng
10 tháng 8 2017 lúc 16:27

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

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

\(=\frac{1}{\left(\sqrt{a}-1\right)\sqrt{a}}:\frac{a-1-\left(a-4\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)

\(=\frac{1}{\left(\sqrt{a}-1\right)\sqrt{a}}:\frac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)

\(=\frac{1}{\left(\sqrt{a}-1\right)\sqrt{a}}.\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)

\(=\frac{\sqrt{a}-2}{3\sqrt{a}}\)

Trần Nguyễn
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Bùi Thế Hào
10 tháng 8 2017 lúc 16:22

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

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

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

\(A=\frac{\sqrt{a}-2}{\sqrt{a}}\)