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25 tháng 6 2019

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 ) 

9 tháng 4 2021

ĐỊT MẸ

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) \(x+\sqrt{x}+1>0\Rightarrow G>0\)

\(x+\sqrt{x}+1>0+0+1=1\)

\(\Rightarrow\frac{2}{x+\sqrt{x}+1}< \frac{2}{1}=2\Rightarrow G< 2\)

\(\Rightarrow O< G< 2\)

14 tháng 5 2021

Em gửi ảnh ạ !

14 tháng 5 2021

Em gửi ảnh trên ạ !!!!!