Question

tìm x,y biết \(\frac{2x+1}{5}=\frac{3y-2}{7}=\)\(\frac{2x+3y-1}{6x}\)

Answer 1

TA CÓ: \(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+3y+1-2}{5+7}=\frac{2x+3y-1}{12}\)

\(\Rightarrow\frac{2x+3y-1}{12}=\frac{2x+3y-1}{6x}\)

\(\Rightarrow6x=12\Rightarrow x=2\)

THAY x=2 VÀO \(\frac{2x+1}{5}\)

CÓ : \(\frac{2x+1}{5}=\frac{2.2+1}{5}=\frac{5}{5}=1\)

\(\Rightarrow\frac{3y-2}{7}=1\left(=\frac{2x+1}{5}\right)\)

\(\Rightarrow3y-2=7\)

\(3y=7-2\)

\(3y=5\)

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

VẬY X=2; Y=5\3

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