Ta có:\(\frac{2x-1}{3}=\frac{x+2}{5}+1\)
\(\frac{2x-1}{3}=\frac{x+7}{5}\)
\(\Rightarrow5\left(2x-1\right)=3\left(x+7\right)\)
\(\Rightarrow10x-5=3x+21\)
\(\Rightarrow10x-3x=21+5\)
\(\Rightarrow7x=26\)
\(\Rightarrow x=\frac{26}{7}\)
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\(\frac{2x-1}{3}=\frac{x+2}{5}+1\)
=> \(\frac{5.\left(2x-1\right)}{15}=\frac{3.\left(x+2\right)}{15}+\frac{15}{15}\)
=> \(5.\left(2x-1\right)=3.\left(x+2\right)+15\)
=> \(10x-5=3x+6+15\)
=> \(10x-3x=5+6+15\)
=> \(7x=26\)
=> \(x=\frac{26}{7}\)
