Question

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Answer 1

$\frac{3x(2-x)^{-1}}{4-(\frac{2-x}{3x})^{-1} } =\frac{3x*\frac{1}{2-x} }{4-\frac{3x}{2-x} } =\frac{3x}{2-x} :(\frac{4(2-x)}{2-x} -\frac{3x}{2-x} )=\frac{3x}{2-x} :\frac{8-4x-3x}{2-x}=\frac{3x}{2-x} *\frac{2-x}{8-7x}=\\\\\frac{3x}{8-7x}$

$\frac{3*\frac{3}{5} }{8-7*\frac{3}{5} }=\frac{9}{5} :(8-\frac{21}{5} )=\frac{9}{5} :\frac{19}{5} =\frac{9*5}{5*19} =\frac{9}{19}$

Answer 2

$\frac{3x(2-x)^{-1} }{4-(\frac{2-x}{3x})^{-1}}=\frac{3x*\frac{1}{2-x}}{4-\frac{3x}{2-x}}=\frac{\frac{3x}{2-x}}{\frac{8-4x-3x}{2-x}}=\frac{3x(2-x)}{(2-x)(8-7x)}=\frac{3x}{8-7x}\\\\x=\frac{3}{5}\\\\\frac{3x}{8-7x}=\frac{3*\frac{3}{5}}{8-7*\frac{3}{5}}=\frac{\frac{9}{5}}{8-\frac{21}{5}}=\frac{9}{5}:\frac{19}{5}=\frac{9}{5}*\frac{5}{19}=\frac{9}{19}\\\\Otvet:\boxed{\frac{3x(2-x)^{-1}}{4-(\frac{2-x}{3x})^{-1}}=\frac{9}{19}}$