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\(\sqrt{\frac{1-x}{x}}=\frac{2x+x^2}{1+x^2}\)
\(\Leftrightarrow\sqrt{\frac{1-x}{x}}-1=\frac{2x+x^2}{1+x^2}-1\)
\(\Leftrightarrow\frac{-\left(2x-1\right)}{\sqrt{\frac{1-x}{x}}+1}-\frac{2x-1}{1+x^2}=0\)
\(\Leftrightarrow\left(2x-1\right)\left(\frac{-1}{\sqrt{\frac{1-x}{x}}+1}-\frac{1}{1+x^2}\right)=0\)
Dễ thấy: \(\frac{-1}{\sqrt{\frac{1-x}{x}}+1}-\frac{1}{1+x^2}< 0\)
\(\Rightarrow2x-1=0\Rightarrow x=\frac{1}{2}\)
\(\left(\frac{4}{27}+\frac{4}{165}+\frac{4}{285}\right):\left(\frac{5}{84}+\frac{3}{180}+\frac{4}{285}\right)=\frac{4}{27}+\frac{4}{165}+\frac{4}{285}:\frac{5}{84}+\frac{3}{180}+\frac{4}{285}=\frac{1052}{5643}:\frac{12}{133}=\frac{1841}{891}\)
\(\frac{2x}{x+1}+\frac{18}{x^2+2x-3}=\frac{2x-5}{x+3}ĐKXĐ:x\ne-1;-3\)
\(\frac{2x}{x+1}+\frac{18}{\left(x-1\right)\left(x+3\right)}=\frac{2x-5}{x+3}\)
\(2x\left(x-1\right)\left(x+3\right)+18\left(x+1\right)=\left(2x-5\right)\left(x+1\right)\left(x-1\right)\)
\(4x^2+12x+18=-2x-5x^2+5\)
\(4x^2+12x+18+2x+5x^2-5=0\)
\(9x^2-14x+13=0\)
=> vô nghiệm