Question

[nguyenthuyhang]

BÀI 2: tìm số hữu tỹ X , biết rằng

a,\(\frac{-3}{2}-2X+\frac{3}{4}=-2\)2

b, (\(\frac{-2}{3}x-\frac{3}{5}\))(\(\frac{3}{-2}-\frac{10}{3}\))=\(\frac{2}{5}\)

c,\(\frac{x}{2}-\left(\frac{3x}{5}-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}.x\right)\)

d,\(\frac{2x-3}{3}+\frac{-3}{2}=\frac{5-3x}{6}-\frac{1}{3}\)

e,\(\frac{2}{3x}-\frac{3}{12}=\frac{4}{5}-\left(\frac{7}{x}-2\right)\)

Answer 1

[Yeji]

a. \(\frac{-3}{2}-2x+\frac{3}{4}=-22\)2

=> \(-2x=-22+\frac{3}{2}-\frac{3}{4}\)

=> \(-2x=\frac{-85}{4}\)

=> \(x=\frac{-85}{4}:\left(-2\right)\)

=> \(x=\frac{85}{8}\)

b. \(\left(\frac{-2}{3}x-\frac{3}{5}\right).\left(\frac{3}{-2}-\frac{10}{3}\right)=\frac{2}{5}\)

=> \(\left(\frac{-2}{3}x-\frac{3}{5}\right).\frac{-29}{6}=\frac{2}{5}\)

=> \(\frac{-2}{3}x-\frac{3}{5}=\frac{2}{5}:\left(\frac{-29}{6}\right)\)

=> \(\frac{-2}{3}x-\frac{3}{5}=\frac{-12}{145}\)

=> \(\frac{-2}{3}x=\frac{-12}{145}+\frac{3}{5}\)

=> \(\frac{-2}{3}x=\frac{15}{29}\)

=> x = \(\frac{15}{29}:\frac{-2}{3}\)

=> x = \(\frac{-45}{58}\)