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4 tháng 9 2020

Ta có: \(A=\left(\frac{1}{\sqrt{x-1}}+\frac{1}{\sqrt{x+1}}\right)^2\times\frac{x^2-1}{2}-\sqrt{1-x^2}\)    \(\left(ĐK:x\ge1\right)\)

   \(\Leftrightarrow A=\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x-1}.\sqrt{x+1}}\right)^2\times\frac{x^2-1}{2}-\sqrt{1-x^2}\)

   \(\Leftrightarrow A=\frac{x+1+x-1+2\sqrt{x^2-1}}{x^2-1}\times\frac{x^2-1}{2}-\sqrt{1-x^2}\)

   \(\Leftrightarrow A=\frac{2x+2\sqrt{x^2-1}}{2}-\sqrt{1-x^2}\)

   \(\Leftrightarrow A=x+\sqrt{1-x^2}-\sqrt{1-x^2}\)

   \(\Leftrightarrow A=x\)

Học tốt

4 tháng 9 2020

ĐKXĐ : ...............

\(A=\left(\frac{1}{\sqrt{x-1}}+\frac{1}{\sqrt{x+1}}\right)^2\times\frac{x^2-1}{2}-\sqrt{1-x^2}\)

\(A=\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{\left(x-1\right)\left(x+1\right)}}\right)^2\times\frac{x^2-1}{2}-\sqrt{1-x^2}\)

\(A=\frac{\left(\sqrt{x+1}+\sqrt{x-1}\right)^2}{x^2-1}\times\frac{x^2-1}{2}-\sqrt{1-x^2}\)

\(A=\frac{x+1+2\sqrt{x^2-1}+x-1}{2}-\sqrt{1-x^2}\)

\(A=\frac{2x+2\sqrt{x^2-1}-2\sqrt{1-x^2}}{2}\)

\(A=\frac{2x+2\sqrt{x^2-1}+2\sqrt{x^2-1}}{2}\)

\(A=\frac{2x+4\sqrt{x^2-1}}{2}\)

\(A=x+2\sqrt{x^2-1}\)

24 tháng 7 2017

a, dk \(x\ge0.x\ne1\)

\(\left(\frac{1+\sqrt{x}+1-\sqrt{x}}{2\left(1-x\right)}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)=\(\left(\frac{1}{1-x}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)

 =\(\left(\frac{1+x-x^2-1}{1-x^2}\right)\left(\frac{x+1}{x}\right)=\frac{x\left(1-x\right)\left(x+1\right)}{x\left(1-x\right)\left(1+x\right)}=1\)

phan b,c ban tu lam not nhe dai lam mk ko lam dau  mk co vc ban rui

11 tháng 7 2018

Bài 1:

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

\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)

\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)

\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)

b)   \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)

\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)

\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)

\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)

c)  ĐK:  \(a\ge0;a\ne1\)

  \(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)

\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)

\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)

\(=1-a+a=1\)

AH
Akai Haruma
Giáo viên
10 tháng 7 2020

Lời giải:

Coi yêu cầu đề là rút gọn. Lần sau bạn chú ý viết đầy đủ đề.

ĐK: $x>0; x\neq 1$
Gọi biểu thức đã cho là $P$. Ta có:

\(P=\frac{x-2+\sqrt{x}}{\sqrt{x}(\sqrt{x}+2)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{(\sqrt{x}-1)(\sqrt{x}+2)}{\sqrt{x}(\sqrt{x}+2)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt{x}}\)

8 tháng 8 2016

a/ đkxđ \(\hept{\begin{cases}\sqrt{1+x}-\sqrt{1-x}\ne0\\\sqrt{1-x^2}-1+x\ne0\\x\ne0\end{cases}}va\hept{\begin{cases}1+x>0\\1-x>0\\1-x^2>0\end{cases}va}\sqrt{\frac{1}{x^2}-1}>0\)

\(\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne1\\-1< x< 1\end{cases}}vax>0\)

b  =/\(\left[\frac{\sqrt{1+x}}{\sqrt{1+x}-\sqrt{1-x}}+\frac{1-x}{\sqrt{1-x^2}-1+x}\right].\left[\frac{\sqrt{1-x^2}}{x}-\frac{1}{x}\right]\)=

\(\left[\frac{\sqrt{1+x}}{\sqrt{1+x}-\sqrt{1-x}}+\frac{1-x}{\sqrt{1-x}\left[\sqrt{1+x}-\sqrt{1-x}\right]}\right].\frac{\sqrt{1-x^2}-1}{x}\)=\(\left[\frac{\sqrt{1+x}}{\sqrt{1+x}-\sqrt{1-x}}+\frac{\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right].\frac{\sqrt{1-x^2}-1}{x}\)=\(\frac{\left[\sqrt{1+x}+\sqrt{1-x}\right]\left[\sqrt{1-x^2}-1\right]}{\left[\sqrt{1+x}-\sqrt{1-x}\right].x}\)

c/ khi x=1/2 thi A=\(\frac{\left[\sqrt{1+\frac{1}{2}}+\sqrt{1-\frac{1}{2}}\right]\left[\sqrt{1-\frac{1}{4}}-1\right]}{\left[\sqrt{1+\frac{1}{2}}-\sqrt{1-\frac{1}{2}}\right].\frac{1}{2}}=-1\)

3 tháng 8 2018

a/ đkxđ

√1+x−√1−x≠0
√1−x2−1+x≠0
x≠0

va{

1+x>0
1−x>0
1−x2>0

va√1x2 −1>0

x≠0
x≠1
−1<x<1

vax>0

b  =/[√1+x√1+x−√1−x +1−x√1−x2−1+x ].[√1−x2x −1x ]=

[√1+x√1+x−√1−x +1−x√1−x[√1+x−√1−x] ].√1−x2−1x =[√1+x√1+x−√1−x +√1−x√1+x−√1−x ].√1−x2−1x =[√1+x+√1−x][√1−x2−1][√1+x−√1−x].x 

c/ khi x=1/2 thi A=[√1+12 +√1−12 ][√1−14 −1][√1+12 −√1−12 ].12  =−1