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13 tháng 9 2018

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

\(C=-x\sqrt{x}+x+\sqrt{x}-1\)

\(D=x-\sqrt{x}+1\)

13 tháng 9 2018

có đáp án kĩ hơn không ạ ?

27 tháng 8 2018

a)  ĐK:  \(0< a< 1\)

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

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

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

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

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

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

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

27 tháng 8 2018

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

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

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

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

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

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

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

b)  Xét:  \(Q^3-Q=\left(a-1\right)^3-\left(a-1\right)=\left(a-1\right)^2\left(a-1-1\right)=\left(a-1\right)^2\left(a-2\right)\)

Do  \(a< 1\)=>  \(a-2< 0\) và   \(a-1< 0\) 

nên \(\left(a-1\right)^2\left(a-2\right)< 0\)

=>  \(Q^3-Q< 0\)

<=> \(Q^3< Q\)

7 tháng 8 2019
https://i.imgur.com/3xuKEN9.jpg
7 tháng 8 2019
https://i.imgur.com/JCFXX2s.jpg
NV
25 tháng 5 2019

ĐKXĐ:...

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

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

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

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

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

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

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 

26 tháng 5 2018

B ơi b lấy đề này ở đâu v ạ