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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
c,\(\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{1-a}{\sqrt{1-a^2}-1+a}\right)\left(\sqrt{\frac{1}{a^2}-1}-\frac{1}{a}\right)\)
\(=\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{\sqrt{1-a}.\sqrt{1-a}}{\sqrt{1-a}\left(\sqrt{1+a}-\sqrt{1-a}\right)}\right)\left(\frac{\sqrt{1-a^2}-1}{a}\right)\)
\(=\frac{\left(\sqrt{1+a}+\sqrt{1-a}\right)^2}{\left(1+a\right)-\left(1-a\right)}.\frac{\left(\sqrt{1-a^2}-1\right)}{a}=-1\)
M chỉ làm tiếp thôi nha, ko chép lại đề với đk đâu
a,
\(=\frac{a+2\sqrt{ab}+b-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\)\(\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\frac{a-2\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}-\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\sqrt{a}+\sqrt{b}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}+\sqrt{b}\)
\(=0\)
b,
\(=\left(a-b\right)\left(\sqrt{\frac{a+b}{a-b}}-1\right)\left(a-b\right)\left(\sqrt{\frac{a+b}{a-b}}+1\right)\)
\(=\left(a-b\right)^2\left(\frac{a+b}{a-b}-1\right)\)
\(=\left(a-b\right)^2\cdot\frac{a+b-a+b}{a-b}\)
\(=\left(a-b\right)2b=2ab-2b^2\)
\(x=\frac{1}{2}\left(\sqrt{\frac{1-a}{a}}+\sqrt{\frac{a}{1-a}}\right)=\frac{1}{2}\frac{1-a+a}{\sqrt{\left(1-a\right)a}}=\frac{1}{2\sqrt{\left(1-a\right)a}}\)
\(x^2=\left(\frac{1}{2\sqrt{\left(1-a\right)a}}\right)^2=\frac{1}{4a\left(1-a\right)}\Leftrightarrow x^2-1=\frac{1-4a+4a^2}{4a\left(1-a\right)}=\frac{\left(2a-1\right)^2}{4a\left(1-a\right)}\)
\(\Leftrightarrow\sqrt{x^2-1}=\sqrt{\frac{\left(2a-1\right)^2}{4a\left(1-a\right)}}\Leftrightarrow2a\sqrt{x^2-1}=2a.\frac{2a-1}{2\sqrt{a\left(1-a\right)}}=\sqrt{a}.\frac{2a-1}{\sqrt{1-a}}\).
\(\sqrt{x^2-1}-x=\frac{2a-1}{2\sqrt{a\left(1-a\right)}}-\frac{1}{2\sqrt{\left(1-a\right)a}}=\frac{2a-1+1}{2\sqrt{a\left(1-a\right)}}=\frac{2a}{2\sqrt{a\left(1-a\right)}}=\frac{\sqrt{a}}{\sqrt{1-a}}\)
=> 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