Cho \(Q=\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{1-a}{\sqrt{1-a^2}-1+a}\right)\left(\sqrt{\frac{1}{a^2}-1}-\frac{1}{a}\right)\sqrt{a^2-2a+1}\left(ĐK:0< a< 1\right)\)
a, Rút gọn Q ( câu này viết kq với hướng làm thôi cũng được !)
b, so sánh Q với Q3
Giúp mình với !!! UHUHUHU
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 \(\left(a-1\right)^2\left(a-2\right)< 0\)
=> \(Q^3-Q< 0\)
<=> \(Q^3< Q\)