2) Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{1+2y}{18}=\dfrac{1+6y}{6x}=\dfrac{1+2y+1+6y}{18+6x}=\dfrac{2\left(1+4y\right)}{2\left(9+3x\right)}=\dfrac{1+4y}{9+3x}\)

⇒ \(\dfrac{1+4y}{9+3x}=\dfrac{1+4y}{28}\)

⇒\(9+3x=28\)

⇒\(3x=19\)

⇒\(x=\dfrac{19}{3}\)

bạn thay vào là tìm được y

a, Ta có: \(7y=5z\Leftrightarrow\dfrac{y}{5}=\dfrac{z}{7}\)

Ta lại có: \(\dfrac{x}{3}=\dfrac{y}{4}\Leftrightarrow\dfrac{x}{15}=\dfrac{y}{20}\left(1\right)\)

\(\dfrac{y}{5}=\dfrac{z}{7}\Leftrightarrow\dfrac{y}{20}=\dfrac{z}{28}\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\Leftrightarrow\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}\) và \(2x+3y-z=186\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{186}{62}=3\)

+) \(\dfrac{2x}{30}=3\Rightarrow2x=3.30=90\Rightarrow x=90:2=45\)

+) \(\dfrac{3y}{60}=3\Rightarrow3y=3.60=180\Rightarrow y=180:3=60\)

+) \(\dfrac{z}{28}=3\Rightarrow z=3.28=84\)

Vậy ...

a, 1+2y / 18 = 1+4y / 24 = 1+6y / 6x

Ta có : 1+2y / 18 = 1+6y / 6x = 1+2y + 1+6y / 18 + 6y

= 2+ 8y / 18+6y = 2 (1+4y) / 2( 9 +3y) = 1+4y/9+3y

Ta lại có : 1 + 4y/24 = 1+4y / 9+3y

=> 24=9+3y => 15=3y => y=5

Vậy y=5

Nhớ like

b, 1+3y/12 = 1+5y/5x = 1+7y/4x

Ta có : 1+3y/12 = 1+7y/4x = 1+3y+1+7y / 12 +4x

= 2 + 10y / 12 +4x = 2 (1+5y) / 2 (6+2x) = 1+5y / 6+2x

Ta lại có: 1+5y / 5x = 1+5y / 6+2x

=> 5x = 6+2x => 3x = 6 => x=2

Vậy x =2

Mình sửa lại câu a

1+2y/18 = 1+6y / 6x = 1+2y+1+6y / 18 + 6x = 2 +8y /18+6x

= 2 (1+4y) / 2 (9 +3x) = 1+4y / 9 +3x

Ta lại có: 1+4y/24 = 1+4y/ 9 +3x

=> 24 = 9 +3x => 15= 3x => x =5

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

Áp dụng tính chất của dãy tỉ số bằng nhau , ta có :

\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}=\dfrac{1+2y+1+6y}{18+6x}=\dfrac{2.\left(1+4y\right)}{2.\left(9+3x\right)}=\dfrac{1+4y}{9+3x}\)

⇒\(\dfrac{1+4y}{9+3x}=\dfrac{1+4y}{24}\)

⇒\(9+3x=24\)

⇒\(3x=24-9=15\)

⇒\(x=15:3=5\)

Ta có : 2x+1 /5 = 3y-2/7 = 2x+3y -1 /6x

=> 2x+1+3y-2 / 5+7 = 2x+3y-1 /6x

=> 2x+3y-1 / 12 = 2x+3y-1 / 6x

=> 12 = 6x => x =2

Ta có:\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}=\frac{2\left(1+2y\right)-\left(1+4y\right)}{2.18-24}=\frac{1+2y+1+4y-\left(1+6y\right)}{18+24-6x}\)

\(\Rightarrow\frac{1}{6}=\frac{1}{6x}\Rightarrow x=1\)

x = 1 nha bạn

k tui nha

thanks

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

=> \(\frac{1+4y}{24}=\frac{1+2y+1+6y}{18+6x}\)=> \(\frac{1+4y}{24}=\frac{2+8y}{2\left[9+3x\right]}\)

=> 9 + 3x = 24 => 3x = 15 => x = 5

Áp dụng tính chất dãy tỉ số bằng nhau,ta có:

\(\dfrac{1+2y}{18}=\dfrac{1+6y}{6x}\Rightarrow\dfrac{1+2y+1+6y}{18+6x}\Rightarrow\dfrac{8y+2}{18+6x}\Leftrightarrow\dfrac{2\left(1+4y\right)}{2\left(9+3x\right)}=\dfrac{1+4y}{9+3x}\)

\(\Rightarrow\dfrac{1+4y}{9+3x}=\dfrac{1+4y}{24}\Leftrightarrow9+3x=24\Rightarrow x=\dfrac{24-9}{3}=5\)

Thay x=5 vào biểu thức: \(\dfrac{1+2y}{18}=\dfrac{1+6y}{6x}\),ta đc:

\(\dfrac{1+2y}{18}=\dfrac{1+6y}{6.5}\Leftrightarrow\dfrac{1+2y}{18}=\dfrac{1+6y}{30}\Leftrightarrow\dfrac{5\left(1+2y\right)}{90}=\dfrac{3\left(1+6y\right)}{90}\)

\(\Leftrightarrow5\left(1+2y\right)=3\left(1+6y\right)\Leftrightarrow5+10y=3+18y\)

\(\Leftrightarrow10y-18y=3-5\Leftrightarrow-8y=-2\Leftrightarrow y=\dfrac{1}{4}=0,25\)

vậy x=5 và y=0,25