Question

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

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

Answer 3

Nếu thấy to thì bạn nhấn tổ hợp phím (Ctrl -) để thấy rõ nhé

Answer 4

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

<=> \(A=\frac{\sqrt{2}\left(1+\sqrt{5}\right)}{2+\sqrt{6+2\sqrt{5}}}+\frac{\sqrt{2}\left(1-\sqrt{5}\right)}{2-\sqrt{6-2\sqrt{5}}}\)

<=> \(A=\frac{\sqrt{2}\left(1+\sqrt{5}\right)}{2+\sqrt{\left(\sqrt{5}+1\right)^2}}+\frac{\sqrt{2}\left(1-\sqrt{5}\right)}{2-\sqrt{\left(\sqrt{5}-1\right)^2}}\)

<=> \(A=\frac{\sqrt{2}\left(1+\sqrt{5}\right)}{2+\sqrt{5}+1}+\frac{\sqrt{2}\left(1-\sqrt{5}\right)}{2-\left|\sqrt{5}-1\right|}\)

<=> \(A=\frac{\sqrt{2}\left(1+\sqrt{5}\right)}{3+\sqrt{5}}+\frac{\sqrt{2}\left(1-\sqrt{5}\right)}{2-\sqrt{5}+1}=\frac{\sqrt{2}\left(1+\sqrt{5}\right)}{3+\sqrt{5}}+\frac{\sqrt{2}\left(1-\sqrt{5}\right)}{3-\sqrt{5}}=\frac{\sqrt{2}\left(1+\sqrt{5}\right)\left(3-\sqrt{5}\right)+\sqrt{2}\left(1-\sqrt{5}\right)\left(3+\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)

<=>\(A=\frac{\sqrt{2}\left(3-\sqrt{5}+3\sqrt{5}-5\right)+\sqrt{2}\left(3+\sqrt{5}-3\sqrt{5}-5\right)}{9-5}\)

<=> \(A=\frac{\sqrt{2}\left(3-\sqrt{5}+3\sqrt{5}-5+3+\sqrt{5}-3\sqrt{5}-5\right)}{4}\)

<=> \(A=\frac{\sqrt{2}\left(6-10\right)}{4}=\frac{\sqrt{2}.\left(-4\right)}{4}=-\sqrt{2}\)

Vậy A=\(-\sqrt{2}\)

Answer 5

Mình đăng lại cả bài nhé

Answer 6