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a) ĐK: \(0< a< 1\)
\(Q=\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{1-a}{\sqrt{1-a^2}-1+a}\right)\left(\sqrt{\frac{1}{a^2}-1}-\frac{1}{a}\right)\sqrt{a^2-2a+1}\)
\(=\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{1-a}{\sqrt{1-a}\left(\sqrt{1+a}-\sqrt{1-a}\right)}\right)\left(\frac{\sqrt{1-a^2}}{a}-\frac{1}{a}\right).\sqrt{\left(1-a\right)^2}\)
\(=\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}\right).\frac{\sqrt{1-a^2}-1}{a}.\left(1-a\right)\)
\(=\frac{\left(\sqrt{1+a}+\sqrt{1-a}\right)^2}{\left(\sqrt{1+a}-\sqrt{1-a}\right)\left(\sqrt{1+a}+\sqrt{1-a}\right)}.\frac{\sqrt{1-a^2}-1}{a}.\left(1-a\right)\)
\(=\frac{2+2\sqrt{1-a^2}}{2a}.\frac{\sqrt{1-a^2}-1}{a}.\left(1-a\right)\)
\(=\frac{\sqrt{1-a^2}+1}{a}.\frac{\sqrt{1-a^2}-1}{a}.\left(1-a\right)\)
\(=\frac{-a^2\left(1-a\right)}{a^2}=a-1\)
\(Q=\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{1-a}{\sqrt{1-a^2}-1+a}\right)\left(\sqrt{\frac{1}{a^2}-1}-\frac{1}{a}\right)\sqrt{a^2-2a+1}\)
\(=\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{1-a}{\sqrt{1-a}\left(\sqrt{1+a}-\sqrt{1-a}\right)}\right)\left(\frac{\sqrt{1-a^2}}{a}-\frac{1}{a}\right).\sqrt{\left(1-a\right)^2}\)
\(=\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}\right).\frac{\sqrt{1-a^2}-1}{a}.\left(1-a\right)\)
\(=\frac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}.\frac{\sqrt{1-a^2}-1}{a}.\left(1-a\right)\)
\(=\frac{2+2\sqrt{1-a^2}}{2a}.\frac{\sqrt{1-a^2}-1}{a}.\left(1-a\right)\)
\(=\frac{\sqrt{1-a^2}+1}{a}.\frac{\sqrt{1-a^2}-1}{a}.\left(1-a\right)\)
\(=\frac{-a^2\left(1-a\right)}{a^2}=a-1\)
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 \(\left(a-1\right)^2\left(a-2\right)< 0\)
=> \(Q^3-Q< 0\)
<=> \(Q^3< Q\)
\(A=1+"\frac{2a+\sqrt{a}-1}{1-a}-\frac{2a\sqrt{a}-\sqrt{a}+a}{1-a\sqrt{a}}"\times\frac{a-\sqrt{a}}{2\sqrt{a}-1}=\)
\(A="\frac{1a+\sqrt{a}-1}{1-a}-\frac{1a\sqrt{a}-\sqrt{a}+a}{1-a\sqrt{a}}"\times\frac{a-\sqrt{a}}{1\sqrt{a}-1}\)
P/s: Ko chắc đâu nhé
Ta có: \(1+\left(\dfrac{2a+\sqrt{a}-1}{1-a}-\dfrac{2a\sqrt{a}-\sqrt{a}+a}{1-a\sqrt{a}}\right)\cdot\dfrac{a-\sqrt{a}}{2\sqrt{a}-1}\)
\(=1+\left(\dfrac{-2\sqrt{a}+1}{\sqrt{a}-1}+\dfrac{\sqrt{a}\left(2\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\right)\cdot\dfrac{a-\sqrt{a}}{2\sqrt{a}-1}\)
\(=1+\left(\dfrac{-\left(2\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)+\sqrt{a}\left(2\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\right)\cdot\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)
\(=1+\dfrac{\left(2\sqrt{a}-1\right)\left(-a-\sqrt{a}-1+a+\sqrt{a}\right)}{a+\sqrt{a}+1}\cdot\dfrac{\sqrt{a}}{2\sqrt{a}-1}\)
\(=1+\dfrac{-\sqrt{a}}{a+\sqrt{a}+1}\)
\(=\dfrac{a+\sqrt{a}+1-\sqrt{a}}{a+\sqrt{a}+1}\)
\(=\dfrac{a+1}{a+\sqrt{a}+1}\)
DKXD a>0
ở tử có chứa mẫu
\(a^2+\sqrt{a}=\sqrt{a}.\left(\sqrt{a^3}+1\right)=\sqrt{a}.\left(\sqrt{a}+1\right)\left(a-\sqrt{a}11\right)\)
cái kia cũng thế
\(A=\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\frac{2a+\sqrt{a}}{\sqrt{a}}+1\left(ĐK:x>0\right)\)
\(=\frac{a^2+\sqrt{a}}{\left(a-\sqrt{a}+1\right)}-2\sqrt{a}-1+1\)
\(=\frac{a^2+\sqrt{a}-2a\sqrt{a}+2a-2\sqrt{a}}{a-\sqrt{a}+1}\)
\(=\frac{a^2-2a\sqrt{a}+2a-\sqrt{a}}{a-\sqrt{a}+1}=\frac{\sqrt{a}.\left(\sqrt{a}-1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}=\sqrt{a}\left(\sqrt{a}-1\right)\)