Question

Tìm x

\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)

Answer 1

Ta có :

\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}\)

\(\Leftrightarrow24\left(1+2y\right)=18\left(1+4y\right)\)

\(\Leftrightarrow24+48y=18.72y\)

\(\Leftrightarrow24-18=72y-48y\)

\(\Leftrightarrow6=24y\)

\(\Leftrightarrow y=\dfrac{1}{4}\)

Thay \(y=\dfrac{1}{4}\) ta có :

\(\dfrac{1+1}{24}=\dfrac{1+\dfrac{3}{2}}{6x}\)

\(=\dfrac{1}{12}=\dfrac{\dfrac{5}{2}}{6x}\)

\(\Leftrightarrow6x=\dfrac{5}{2}.12\)

\(\Leftrightarrow6x=30\)

\(\Leftrightarrow x=5\)

Answer 2

*ta có:

\(\dfrac{1+2y}{18}\)=\(\dfrac{1+4y}{24}\)

<=>(1+2y).24=18.(1+4y)

<=>24+48y =18+72y

<=>24-18 = 72y-48y

<=>6 = 24y

<=> y =\(\dfrac{1}{4}\)

* Thay y=\(\dfrac{1}{4}\)vào tỉ lệ thức\(\dfrac{1+4y}{24}\)=\(\dfrac{1+6y}{6x}\),ta có:

\(\dfrac{1+4^{\dfrac{1}{4}}}{18}\)\(\)=\(\dfrac{1+6\dfrac{1}{4}}{6x}\)

<=>\(\dfrac{1+1}{24}\)=\(\dfrac{1+\dfrac{3}{2}}{6x}\)

<=>\(\dfrac{1}{12}\)=\(\dfrac{\dfrac{5}{2}}{6x}\)

<=>\(\dfrac{1}{12}\)=\(\dfrac{5}{12x}\)

<=>12.5=12x

<=>x=5

Vậy x=5;y=\(\dfrac{1}{4}\)