\(\dfrac{x}{4}=\dfrac{2y+1}{3}=\dfrac{x-2y-1}{y}=\dfrac{x-2y-1-x+2y+1}{4-3-y}=\dfrac{0}{1-y}=0\\ \Rightarrow\left\{{}\begin{matrix}x=0\\2y+1=0\\x-2y-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\y=-\dfrac{1}{2}\end{matrix}\right.\)

Áp dụng t/c dtsbn ta có:

\(\dfrac{x}{4}=\dfrac{2y+1}{3}=\dfrac{x-2y-1}{y}=\dfrac{x-2y-1}{4-3}=\dfrac{x-2y-1}{1}=x-2y-1\)

\(\dfrac{x-2y-1}{y}=x-2y-1\Rightarrow x-2y-1=y\left(x-2y-1\right)\Rightarrow\left(y-1\right)\left(x-2y-1\right)=0\Rightarrow\left[{}\begin{matrix}y=1\\x-2y-1=0\end{matrix}\right.\)

Với y=1:\(\dfrac{x}{4}=\dfrac{2y+1}{3}=\dfrac{2.1+1}{3}=1\Rightarrow x=4\)

Với \(x-2y-1=0\)\(\Rightarrow\dfrac{x}{4}=\dfrac{2y+1}{3}=0\Rightarrow\left\{{}\begin{matrix}x=0\\y=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy \(\left(x,y\right)\in\left\{\left(4;1\right);\left(0;-\dfrac{1}{2}\right)\right\}\)

cảm ơn  2 anh ạ

A ) ĐK: x#0 
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)

B ) x+y=3(x−y)=x:y
→x+y=3x−3y
→4y=2x
→x:y=4:2=2
→x+y=2
Mà x=2y nên
2y+y=3y=2
→y=2/3
→x=2−2/3=4/3

Chú ý : dấu / nghĩa là phần

Nếu mình đúng thì các bạn k mình nhé

a) \(\frac{1+2y}{18}=\frac{1+4y}{24}\Rightarrow24+48y=18+72y\Rightarrow6=24y\Rightarrow y=\frac{1}{4}\)

\(\frac{1+4y}{24}=\frac{1+6y}{6x}\Rightarrow\frac{1+4.\frac{1}{4}}{24}=\frac{1+6.\frac{1}{4}}{6x}\Rightarrow\frac{2}{24}=\frac{\frac{5}{2}}{6x}\Rightarrow12x=60\Rightarrow x=5\)

b) \(x+y=3\left(x-y\right)\Rightarrow x+y=3x-3y\Rightarrow4y=2x\Rightarrow x=2y\)

\(x+y=\frac{x}{y}\Rightarrow2y+y=\frac{2y}{y}\Rightarrow3y=2\Rightarrow y=\frac{2}{3}\Rightarrow x=2y=\frac{4}{3}\)

\(\)

a) \(\frac{1+2y}{18}=\frac{1+4y}{24}\) => (1+2y).24=18.(1+4y) => 24+48y=18+72y => y=\(\frac{1}{4}\)

=> \(\frac{1+6y}{6x}=\frac{1+6\cdot\frac{1}{4}}{6x}=\frac{\frac{5}{2}}{6x}=\frac{1+2\cdot\frac{1}{4}}{18}=\frac{1}{12}\)=> x= \(\frac{\frac{5}{2}\cdot12}{6}=5\)

b) x+y=3.(x-y) => x+y=3x-3y =>2x=4y => x/y=2 => x+y=3.(x-y)=2 => x-y=2/3

=> x=4/3; y=2/3

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

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

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

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

\(\Rightarrow\)6=24y

\(\Rightarrow\)y=\(\frac{6}{24}\)

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

Thay y=\(\frac{1}{4}\) vào đề ta có:

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

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

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

=>30=6x

=>x=5

Vậy x=5;y=\(\frac{1}{4}\)

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

Ta có =

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

24+48y=18+72y

24-18=72y-48y

6=24y

y=

y=

Thay y= vào đề ta có:

1 + 2 / 18 = 1 + 4 / 24 = 1 + 6 / 6x

=>=

=>12.=6x

=>30=6x

=>x=5

Vậy x=5;y=

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

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

(2x + y) + (2y - x)i = (x - 2y + 3) + (y + 2x + 1)i