\(\frac{-15}{12}x+\frac{3}{7}=\frac{6}{5}x-\frac{1}{2}\\ \frac{-5}{4}x+\frac{3}{7}=\frac{6}{5}x-\frac{1}{2}\\ \frac{6}{5}x+\frac{5}{4}x=\frac{1}{2}+\frac{3}{7}\\ x\left(\frac{6}{5}+\frac{5}{4}\right)=\frac{7}{14}+\frac{6}{14}\\ x\left(\frac{24}{20}+\frac{25}{20}\right)=\frac{13}{14}\\ x\cdot\frac{49}{20}=\frac{13}{14}\\ x=\frac{13}{14}\div\frac{49}{20}\\ x=\frac{13}{14}\cdot\frac{20}{49}\\ x=\frac{130}{343}\\ \text{Vậy }x=\frac{130}{343}\)
\(-\frac{15}{12}x+\frac{3}{7}=\frac{6}{5}x-\frac{1}{2}\)
\(\Rightarrow\left(-\frac{15}{12}x\right)-\frac{6}{5}x=\left(-\frac{1}{2}\right)-\frac{3}{7}\)
\(\Rightarrow-\frac{49}{20}x=-\frac{13}{14}\)
\(\Rightarrow\frac{49}{20}x=\frac{13}{14}\)
\(\Rightarrow x=\frac{13}{14}:\frac{49}{20}\)
\(\Rightarrow x=\frac{130}{343}\)
Vậy \(x=\frac{130}{343}.\)
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