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\(x=\frac{1}{2}\left(\sqrt{\frac{1-a}{a}}+\sqrt{\frac{a}{1-a}}\right)=\frac{1}{2}\frac{1-a+a}{\sqrt{\left(1-a\right)a}}=\frac{1}{2\sqrt{\left(1-a\right)a}}\)
\(x^2=\left(\frac{1}{2\sqrt{\left(1-a\right)a}}\right)^2=\frac{1}{4a\left(1-a\right)}\Leftrightarrow x^2-1=\frac{1-4a+4a^2}{4a\left(1-a\right)}=\frac{\left(2a-1\right)^2}{4a\left(1-a\right)}\)
\(\Leftrightarrow\sqrt{x^2-1}=\sqrt{\frac{\left(2a-1\right)^2}{4a\left(1-a\right)}}\Leftrightarrow2a\sqrt{x^2-1}=2a.\frac{2a-1}{2\sqrt{a\left(1-a\right)}}=\sqrt{a}.\frac{2a-1}{\sqrt{1-a}}\).
\(\sqrt{x^2-1}-x=\frac{2a-1}{2\sqrt{a\left(1-a\right)}}-\frac{1}{2\sqrt{\left(1-a\right)a}}=\frac{2a-1+1}{2\sqrt{a\left(1-a\right)}}=\frac{2a}{2\sqrt{a\left(1-a\right)}}=\frac{\sqrt{a}}{\sqrt{1-a}}\)
=> B=\(\frac{2a\sqrt{X^2-1}}{x-\sqrt{x^2-1}}=\frac{\sqrt{a}\left(2a-1\right)}{\sqrt{1-a}}:\frac{\sqrt{a}}{\sqrt{1-a}}=\frac{\sqrt{a}\left(2a-1\right)}{\sqrt{1-a}}.\frac{\sqrt{1-a}}{\sqrt{a}}=2a-1\)
mình chỉ rút gọn được ghi đề là: \(x^2-1\) thôi. nếu như đề của bạn thì: \(x^2=\left(\frac{1}{2\sqrt{\left(1-a\right)a}}\right)^2=\frac{1}{4a\left(1-a\right)}\Leftrightarrow x^2+1=\frac{1+4a-4a^2}{4a\left(1-a\right)}=\)mình k thể rút gọn được nữa đâu
c,Có x=\(\frac{1}{2}\left(\sqrt{\frac{1-a}{a}}-\sqrt{\frac{a}{1-a}}\right)\left(0< a< 1\right)\)
<=> \(x=\frac{1}{2}\left(\frac{\sqrt{1-a}}{\sqrt{a}}-\frac{\sqrt{a}}{\sqrt{1-a}}\right)\) (vì 0<a<1)
<=>\(x=\frac{1}{2}.\frac{\sqrt{1-a}^2-\sqrt{a}^2}{\sqrt{a}.\sqrt{1-a}}=\frac{1}{2}.\frac{1-a-a}{\sqrt{a\left(1-a\right)}}=\frac{1}{2}.\frac{1-2a}{\sqrt{a\left(1-a\right)}}=\frac{1-2a}{2\sqrt{a\left(1-a\right)}}\)(1)
<=> 1+x2=1+\(\frac{1}{4}.\frac{\left(1-2a\right)^2}{a\left(1-a\right)}\)= \(\frac{4a\left(1-a\right)+\left(1-2a\right)^2}{4a\left(1-a\right)}\)
<=> 1+x2=\(\frac{4a-4a^2+1-4a+4a^2}{4a\left(1-a\right)}=\frac{1}{4a\left(1-a\right)}\)>0
<=> \(\sqrt{1+x^2}=\frac{1}{2\sqrt{a\left(1-a\right)}}\) (2)
Thay (1),(2) vào C có:
C= \(\frac{2a.\frac{1}{2\sqrt{a\left(1-a\right)}}}{\frac{1}{2\sqrt{a\left(1-a\right)}}-\frac{1-2a}{2\sqrt{a\left(1-a\right)}}}=\frac{\frac{a}{\sqrt{a\left(1-a\right)}}}{\frac{1-1+2a}{2\sqrt{a\left(1-a\right)}}}=\frac{\frac{a}{\sqrt{a\left(1-a\right)}}}{\frac{2a}{2\sqrt{a\left(1-a\right)}}}=1\)
Vậy C=1
