Question

\(a,\frac{1}{3}x+\frac{2}{5}\left(x-1\right)=0\)

\(b,\left(2x-3\right).\left(6-2x\right)=0\)

\(c,x:\frac{3}{4}+\frac{1}{4}=-\frac{2}{3}\)

\(d,\frac{-2}{3}-\frac{1}{3}.\left(2x-5\right)=\frac{3}{2}\)

\(60\%x+\frac{2}{3}x=\frac{1}{3}.6\frac{1}{3}\)

Answer 1

a)\(\frac{1}{3}x+\frac{2}{5}\left(x-1\right)=0\)

\(\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)

\(\frac{11}{15}x-\frac{2}{5}=0\)

\(\frac{11}{15}x=\frac{2}{5}\)

\(x=\frac{6}{11}\)

b)(2x-3)(6-2x)=0

=>2x-3=0 hoặc 6-2x=0

=>x=3/2 hoặc x=3

c)\(x:\frac{3}{4}+\frac{1}{4}=-\frac{2}{3}\)

\(x:\frac{3}{4}=-\frac{11}{12}\)

\(x=-\frac{11}{16}\)

d)\(-\frac{2}{3}-\frac{1}{3}\left(2x-5\right)=\frac{3}{2}\)

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

\(-\frac{2}{3}x+1=\frac{3}{2}\)

\(-\frac{2}{3}x=\frac{1}{2}\)

\(x=-\frac{3}{4}\)

\(60\%x+\frac{2}{3}x=\frac{1}{3}.6\frac{1}{3}\)

\(\frac{3}{5}x+\frac{2}{3}x=\frac{19}{9}\)

\(\frac{19}{15}x=\frac{19}{9}\)

\(x=\frac{5}{3}\)