Ta có: 
(1) 1+2y/18 = 1+4y/24 
=> 24 + 48y = 18 + 72y 
<=> y=1/4 
(2) 1+4y/24=1+6y/6x 
Thay y=1/4 vào (2) ta tìm đc x=5 (thỏa

cho 3 tỉ số = nhau là : a/b+c; b/c+a; c/a+b

tìm gt của mỗi tỉ số đó

ta co : 1+2y/18=1+4y/24

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

=>24+48y=18+72y

=>24-18=72y-48y

=>6=24y

=>y=1/4 

thay y thanh 1/4 vao de bai ta co : 

1+1/2/18=1+1/24=(1+3/2)/6x

=>1/12=(5/2)/6x

=>12/(5/2)=6x

=>30=6x/x=5

vay x=5 va y=1/4

\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)(ĐK: \(x\ne0\))
\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}\)
\(\Rightarrow\left(1+2y\right)24=\left(1+4y\right)18\)
\(\Rightarrow24+48y=18+72y\)
\(\Rightarrow72y-48y=24-18\)
\(\Rightarrow24y=6\)
\(\Rightarrow y=\dfrac{1}{4}\) \(\left(1\right)\)
Ta có: \(\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\) \(\left(2\right)\)
Thay \(\left(1\right)\) vào \(\left(2\right)\), ta có:
\(\dfrac{1+4\cdot\dfrac{1}{4}}{24}=\dfrac{1+6\cdot\dfrac{1}{4}}{6x}\)
\(\Rightarrow\dfrac{2}{24}=\dfrac{\dfrac{5}{2}}{6x}\)
\(\Rightarrow6x=\dfrac{\dfrac{5}{2}\cdot24}{2}\)
\(\Rightarrow6x=30\)
\(\Rightarrow x=5\)(thỏa mãn)
Vậy x = 5 và y = \(\dfrac{1}{4}\)
#YM

\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}=\frac{\left(1+2y\right)+\left(1+6y\right)}{18+6x}\)(tính chất dãy tỉ số bằng nhau)

\(=\frac{2+8y}{2\left(9+3x\right)}\)

\(=\frac{2\left(1+4y\right)}{2\left(9+3x\right)}=\frac{1+4y}{9+3x}\)

\(\Rightarrow\frac{1+4y}{24}=\frac{1+4y}{9+3x}\Rightarrow9+3x=24\Rightarrow x=\frac{\left(24-9\right)}{3}=5\)

Vậy x = 5

x = 5

k cho mk nha

ai k mk mk k lại

Ý bạn là :\(1+\frac{2y}{18}=1+\frac{4y}{24}=1+\frac{6y}{6x}\)

hay \(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\)ạ ??

Lần sau ghi rõ :>

Ta có: \(\frac{1+2y}{18}=\frac{1+4y}{24}\)

=> \(\left(1+2y\right).24=\left(1+4y\right).18\)

=> \(24+48y=18+72y\)

=> \(24-18=72y-48y\)

=> \(24y=6\)

=> \(y=\frac{1}{4}\)

Với y = 1/4 => \(\frac{1+4\cdot\frac{1}{4}}{24}=\frac{1+6\cdot\frac{1}{4}}{6x}\)

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

=> \(6x=\frac{5}{2}.12\)

=> \(6x=30\)

=> \(x=5\)

(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

(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 !

\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}=\frac{1+2y+1+6y}{18+6x}=\frac{2+8y}{6\left(3+x\right)}=\frac{1+4y}{3\left(3+x\right)}\)

\(\Rightarrow3\left(3+x\right)=24\)\(\Rightarrow3+x=8\)\(\Rightarrow x=5\)

Vậy \(x=5\)

Ta có: \(\frac{1+2y}{18}=\frac{1+4y}{24}\)

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

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

\(\Leftrightarrow24y-6=0\Leftrightarrow y=\frac{1}{4}\)

\(\Rightarrow\frac{1+2y}{18}=\frac{1+6y}{6x}\Leftrightarrow\frac{1+\frac{1}{2}}{18}=\frac{1+\frac{3}{2}}{6x}\)

\(\Leftrightarrow x=5\)

Vậy x = 5 và \(y=\frac{1}{4}\)

\(\Rightarrow\)\(\frac{1+2y+1+4y+1+6y}{18+24+6x}\)=\(\frac{\left(1+1+1\right)+2y+4y+6y}{6\left(3+4+x\right)}=\frac{y\left(2+4+6\right)+3}{6\left(3+4+x\right)}=\frac{3+y.12}{6\left(7+x\right)}\)

=\(\frac{3\left(1+4y\right)}{3.2\left(7+x\right)}=\frac{1+4y}{14+2x}\)

\(\Rightarrow\)\(\frac{1}{14}=\frac{2y}{x}\Rightarrow x=14.2y=28y\)

\(\frac{x}{y}=28\)

Ta có: \(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}=\frac{1+2y+1+6y}{18+6x}=\frac{2\left(1+4y\right)}{6\left(x+3\right)}=\frac{1+4y}{3x+9}\)

\(=>\frac{1+4y}{24}=\frac{1+4y}{3x+9}\)\(=>3x+9=24\)

<=>3x=15

<=>x=5

Vậy x có giá trị bằng 5

Chúc bạn học tốt!