Question

[Ánh Right]

Rút gọn các biểu thức sau:

a) A=\(\frac{1+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{1-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)

b) B=\(\left(\frac{1}{\sqrt{x-1}}+\frac{1}{\sqrt{x+1}}\right):\left(\frac{1}{\sqrt{x-1}}-\frac{1}{\sqrt{x+1}}\right)\)

c) C= \(\frac{2a\sqrt{1+x^2}}{\sqrt{1+x^2}-x}\)với \(x=\frac{1}{2}\left(\sqrt{\frac{1-a}{a}}-\sqrt{\frac{a}{1-a}}\right)\)và 0<a<1

Mọi người giải chi tiết giúp mk với ạ

Answer 1

[Lê Thị Thục Hiền]

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

Answer 2

[Kudo Shinichi]