Cho \(\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}=\sqrt{2}\)
Chứng minh rằng \(\frac{x-1}{x+1}=12\sqrt{2}-17\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
VH
10 tháng 10 2019
a, \(\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)
\(=\frac{2+\sqrt{3}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{2-\sqrt{3}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\frac{2+\sqrt{3}}{2+\sqrt{3}+1}+\frac{2-\sqrt{3}}{2-\sqrt{3}+1}\)
\(=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(=\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)
\(=\frac{6+\sqrt{3}-3+6-\sqrt{3}-3}{9-3}=\frac{6}{6}=1\)
b, \(\frac{1}{x+\sqrt{x}}+\frac{2\sqrt{x}}{x-1}-\frac{1}{x-\sqrt{x}}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x-1}\right)}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}-1+2x-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{2\left(x-1\right)}{\sqrt{x}\left(x-1\right)}=\frac{2}{\sqrt{x}}\)
21 tháng 9 2020
a)\(G=\left(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right):\frac{\sqrt{x}-1}{2}\)
\(=\frac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2}{\sqrt{x}-1}\)
\(=\frac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2}{\sqrt{x}-1}\)
\(=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2}{\sqrt{x}-1}\)
\(=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2}{\sqrt{x}-1}\)
\(=\frac{2}{x+\sqrt{x}+1}\)
b) \(x+\sqrt{x}+1>0\Rightarrow G>0\)
\(x+\sqrt{x}+1>0+0+1=1\)
\(\Rightarrow\frac{2}{x+\sqrt{x}+1}< \frac{2}{1}=2\Rightarrow G< 2\)
\(\Rightarrow O< G< 2\)
ĐK: \(-1\le x\le1\)\(;\)\(x\ne0\)
\(\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}=\sqrt{2}\)
\(\Leftrightarrow\)\(\frac{\left(\sqrt{1+x}+\sqrt{1-x}\right)^2}{\left(\sqrt{1+x}-\sqrt{1-x}\right)\left(\sqrt{1+x}+\sqrt{1-x}\right)}=\sqrt{2}\)
\(\Leftrightarrow\)\(\frac{1+x+1-x+2\sqrt{\left(1+x\right)\left(1-x\right)}}{1+x-1+x}=\sqrt{2}\)
\(\Leftrightarrow\)\(\sqrt{1-x^2}=\sqrt{2}x-1\)
\(\Leftrightarrow\)\(1-x^2=2x^2-2\sqrt{2}x+1\)
\(\Leftrightarrow\)\(x^2-\frac{2\sqrt{2}}{3}x=0\)
\(\Leftrightarrow\)\(x\left(x-\frac{2\sqrt{2}}{3}\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\left(l\right)\\x=\frac{2\sqrt{2}}{3}\left(tm\right)\end{cases}}\)
\(\frac{x-1}{x+1}=\frac{\frac{2\sqrt{2}}{3}-1}{\frac{2\sqrt{2}}{3}+1}=\frac{\frac{2\sqrt{2}-3}{3}}{\frac{2\sqrt{2}+3}{3}}=\frac{2\sqrt{2}-3}{2\sqrt{2}+3}=12\sqrt{2}-17\) ( giống như tìm x ở trên )