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

Ta có: 

\(\frac{1+2y}{18}=\frac{1+4y}{24}\Leftrightarrow24+48y=18+72y\Leftrightarrow y=\frac{1}{4}\Rightarrow\frac{1+4y}{24}=\frac{1}{12}\).

\(\frac{1+6y}{6x}=\frac{1}{12}\Leftrightarrow x=2\left(1+6y\right)=2\left(1+\frac{3}{2}\right)=5\).

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

Ta có :

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

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

\(\Rightarrow\frac{1+4y}{24}=\frac{2+8y}{2\left(9+3x\right)}\)

\(\Rightarrow9+3x=24\)

\(3x=15\)

\(\Rightarrow x=5\)

\(\left\{{}\begin{matrix}\dfrac{2y+1}{18}=\dfrac{4y+1}{24}\\\dfrac{4y+1}{24}=\dfrac{6y+1}{6x}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4\left(2y+1\right)=3\left(4y+1\right)\\\dfrac{4y+1}{4}=\dfrac{6y+1}{x}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}12y+3-8y-4=0\\\dfrac{4y+1}{4}=\dfrac{6y+1}{x}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{4}\\\dfrac{2}{4}=\dfrac{\dfrac{3}{2}+1}{x}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{4}\\\dfrac{5}{2}:x=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left(x,y\right)=\left(5;\dfrac{1}{4}\right)\)

(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 \(\left\{{}\begin{matrix}x=5\\y=\dfrac{1}{4}\end{matrix}\right.\)

xét:\(\frac{1+2y}{18}=\frac{1+6y}{6x}=\frac{1+2y+1+6y}{18+6x}=\frac{2+8y}{18+6x}=\frac{2\left(1+4y\right)}{2\left(9+3x\right)}=\frac{1+4y}{9+3x}\) (t/c dãy tỉ số=nhau)

=>\(\frac{1+4y}{24}=\frac{1+4y}{9+3x}\Rightarrow24=9+3x\Rightarrow x=5\)

vậy x=5 thỏa mãn

Ta có: 1 + 2y/18 = 1 + 4y/24 = 1 + 6y/6x = 1 + 2y + 1 + 6y/18 + 6x = 2 + 8y/18 + 6x = 2(1 + 4y)/18 + 6x 
 \ 18 + 6x = 2 . 24 = 48
 \ 6x = 30
 \ x = 5

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)

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

\(\Rightarrow\) 24 . (1 + 2y) = 18 . (1 + 4y)

\(\Rightarrow\) 24 + 48y = 18 + 72y

\(\Rightarrow\) 24 - 18 = 72y - 48y

\(\Rightarrow\) 6 = 24y

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

Thay y = \(\dfrac{1}{4}\) ta có:

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

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

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

\(\Rightarrow\) \(6x=30\)

\(\Rightarrow\) \(x=5\)

Vậy x = 5 và y = \(\dfrac{1}{4}\)

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

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