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\(a,x+\frac{3}{4}=\frac{1}{4}\)
\(\Rightarrow x=\frac{1}{4}-\frac{3}{4}=-\frac{1}{2}\)
\(b,\frac{2}{3}x-\frac{3}{7}=\frac{1}{7}\)
\(\Rightarrow\frac{2}{3}x=\frac{1}{7}+\frac{3}{7}\)
\(\Rightarrow\frac{2}{3}x=\frac{4}{7}\)
\(\Rightarrow x=\frac{4}{7}:\frac{2}{3}=\frac{4}{7}\cdot\frac{3}{2}=\frac{6}{7}\)
\(c,1\frac{2}{3}x-50\%x=-1\frac{1}{6}\)
\(\Rightarrow\frac{5}{3}x-\frac{50}{100}x=-\frac{7}{6}\)
\(\Rightarrow\left[\frac{5}{3}-\frac{50}{100}\right]x=-\frac{7}{6}\)
\(\Rightarrow\left[\frac{5}{3}-\frac{1}{2}\right]x=-\frac{7}{6}\)
\(\Rightarrow\frac{7}{6}x=-\frac{7}{6}\Leftrightarrow x=-\frac{7}{6}:\frac{7}{6}=-\frac{7}{6}\cdot\frac{6}{7}=-1\)
a,x+\(\frac{3}{4}\)=\(\frac{1}{4}\)
=>x =\(\frac{1}{4}\)-\(\frac{3}{4}\)
=> x =\(\frac{-1}{2}\)
Vậy x=\(\frac{-1}{2}\)
b,\(\frac{2}{3}\)x-\(\frac{3}{7}\)=\(\frac{1}{7}\)
=>\(\frac{2}{3}\)x =\(\frac{1}{7}\)+\(\frac{3}{7}\)
=>\(\frac{2}{3}\)x =\(\frac{4}{7}\)
=> x =\(\frac{4}{7}\) :\(\frac{2}{3}\)
=> x =\(\frac{4}{7}\).\(\frac{3}{2}\)
=> x =\(\frac{6}{7}\)
Vậy x=\(\frac{6}{7}\)
c,\(\frac{12}{3}\)x-50%x =\(\frac{-11}{6}\)
=>\(\frac{12}{3}\)x-\(\frac{1}{2}\)x=\(\frac{-11}{6}\)
=>x(\(\frac{12}{3}\)-\(\frac{1}{2}\))=\(\frac{-11}{6}\)
=>x(\(\frac{24}{6}\)-\(\frac{3}{6}\))=\(\frac{-11}{6}\)
=>x\(\frac{21}{6}\) =\(\frac{-11}{6}\)
=>x =\(\frac{-11}{6}\):\(\frac{21}{6}\)
=>x =\(\frac{-11}{6}\).\(\frac{6}{21}\)
=>x =\(\frac{-11}{21}\)
Vậy x=\(\frac{-11}{21}\)
a)
\(\frac{2}{3}+\frac{1}{3}:x=\frac{3}{5}\)
\(\frac{1}{3}:x=\frac{3}{5}-\frac{2}{3}\)
\(\frac{1}{3}:x=\frac{-1}{15}\)
\(x=\frac{1}{3}:\frac{-1}{15}\)
\(x=-5\)
b)
\(\frac{10}{3x}+\frac{67}{4}=-13.25\)
\(\frac{10}{3x}+\frac{67}{4}=\frac{-53}{4}\)
\(\frac{10}{3x}=\frac{-53}{4}+\frac{67}{4}\)
\(\frac{10}{3x}=\frac{7}{2}\)
\(\Rightarrow10.2=7.3x\)
\(20=21x\)
\(x=\frac{20}{21}\)
c)
x+30%x=-1.3
x+0.3x=-1.3
x (1+0.3) = -1.3
x . 1.3 = -1.3
x = -1.3 : 1.3
x = 1
d)
\(\left(\frac{14}{5x}-50\right):\frac{2}{3}=51\)
\(\frac{14}{5x}-50=51.\frac{2}{3}\)
\(\frac{14}{5x}-50=34\)
\(\frac{14}{5x}=34+50\)
\(\frac{14}{5x}=84\)
\(84.5x=14\)
420x = 14
\(x=\frac{1}{30}\)
a, \(\frac{2}{3}+\frac{1}{3}:x=\frac{3}{5}\)
=> \(\frac{1}{3}:x=\frac{3}{5}-\frac{2}{3}\)
=>\(\frac{1}{3}:x=\frac{-1}{15}\)
=> \(x=\frac{1}{3}:\frac{-1}{15}\)
=> \(x=\frac{-1}{5}\)