đề có đúng như z ko bn:
ta có: \(\frac{1+3y}{15}=\frac{1+6y}{18}\)
\(\Rightarrow\left(1+3y\right).18=\left(1+6y\right).15\)
\(18+54y=15+90y\)
\(54y-90y=15-18\)
\(-36y=-3\)
\(y=-3:-36\)
\(y=\frac{1}{12}\)
ta có: \(\frac{1+6y}{18}=\frac{1+9y}{9x}\)
\(\Leftrightarrow\left(1+6y\right).9x=\left(1+9y\right).18\)
\(9x+54xy=18+162y\)
thay số: \(9x+54.\frac{1}{12}x=18+162.\frac{1}{12}\)
\(9x+\frac{9}{2}x=18+\frac{27}{2}\)
\(x.\left(\frac{9}{2}+9\right)=31\frac{1}{2}\)
\(x.13\frac{1}{2}=31\frac{1}{2}\)
\(x=31\frac{1}{2}:13\frac{1}{2}\)
\(x=45\)
KL: x =45 ; y= 1/12
CHÚC BN HỌC TỐT!!!!