(1+2y)/18=(1+6y)/6x=(1+2y+1+6y)/18+6x=(2+8y)/18+6x=(1+4y)/9+3x
=>(1+4y)/9+3x=(1+4y)/24 =>9+3x=24 =>3x=15 =>x=5
(1+2y)/18=(1+6y)/6x=(1+2y+1+6y)/18+6x=(2+8y)/18+6x=(1+4y)/9+3x
=>(1+4y)/9+3x=(1+4y)/24 =>9+3x=24 =>3x=15 =>x=5
Nhé bạn !
\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)
Ta có :
\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}\)
\(\Leftrightarrow24\left(1+2y\right)=18\left(1+4y\right)\)
\(\Leftrightarrow24+48y=18+72y\)
\(\Leftrightarrow72y-48y=24-18\)
\(\Leftrightarrow24y=6\)
\(\Leftrightarrow y=\dfrac{6}{24}=\dfrac{1}{4}\)
mà \(\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)
\(\Leftrightarrow6x\left(1+4y\right)=24\left(1+6y\right)\)
\(\Leftrightarrow6x\left(1+4.\dfrac{1}{4}\right)=24\left(1+6.\dfrac{1}{4}\right)\left(thay.y=\dfrac{1}{4}.vào\right)\)
\(\Leftrightarrow6x.2=24\left(1+\dfrac{3}{2}\right)\)
\(\Leftrightarrow12x=24.\dfrac{5}{2}\)
\(\Leftrightarrow12x=60\)
\(\Leftrightarrow x=5\)
Vậy \(\left\{{}\begin{matrix}x=5\\y=\dfrac{1}{4}\end{matrix}\right.\)
