\(\frac{1}{1-\sqrt{1-x}}-\frac{1}{1+\sqrt{1+x}}=\frac{\sqrt{3}}{x}\)
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22 tháng 6 2016
c) \(C=\frac{\left(2\sqrt{x}+x\right)\left(\sqrt{x}+1\right)-\left(x\sqrt{x}-1\right)}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{x+\sqrt{x}+1-\left(\sqrt{x}+2\right)}{x+\sqrt{x}+1}=\)
\(C=\frac{x\sqrt{x}+2x+x+2\sqrt{x}-x\sqrt{x}+1}{\left(\left(\sqrt{x}\right)^3-1\right)\left(\sqrt{x}+1\right)}\times\frac{x+\sqrt{x}+1}{x-1}=\)
\(C=\frac{3x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\times\frac{x+\sqrt{x}+1}{x-1}=\)
\(C=\frac{3x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\times\frac{1}{x-1}=\)
\(C=\frac{3x+2\sqrt{x}+1}{x-1}\times\frac{1}{x-1}=\frac{3x+2\sqrt{x}+1}{\left(x-1\right)^2}.\)
bn hok sớm thé
ĐK x\(\le\)-1
\(\frac{1}{1-\sqrt{1-x}}-\frac{1}{1+\sqrt{1+x}}=\frac{\sqrt{3}}{x}\)
\(\Leftrightarrow\frac{1+\sqrt{1+x}-1+\sqrt{1-x}}{x}=\frac{\sqrt{3}}{x}\)
\(\Leftrightarrow\frac{2\sqrt{1+x}}{x}=\frac{\sqrt{3}}{x}\)\(\Leftrightarrow2x\sqrt{1+x}=\sqrt{3}x\)
\(\Leftrightarrow4x^2\left(1+x\right)=3x^2\Leftrightarrow4x^3+x^2=0\)
\(\Leftrightarrow x^2\left(4x+1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\left(KTM\right)\\x=\frac{-1}{4}\left(TM\right)\end{cases}}\)