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

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

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

\(\dfrac{1+2y}{18}=\dfrac{1+6y}{6x}\\ \Rightarrow\dfrac{1+2y}{18}=\dfrac{1+6y}{6x}=\dfrac{\left(1+2y\right)+\left(1+6y\right)}{18+6x}\\ =\dfrac{2+8y}{18+6x}=\dfrac{1+4y}{9+3x}\\ \Rightarrow9+3x=24\\ \Rightarrow3x=15\\ \Rightarrow x=5\)

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

\(\Rightarrow24+48y=18+72y\)

\(\Rightarrow6=24y\Rightarrow y=\dfrac{1}{4}\)

\(\Rightarrow\dfrac{1+\dfrac{1}{2}}{18}+\dfrac{1+\dfrac{3}{2}}{6x}\)

\(\Rightarrow9x=45\Rightarrow x=5\)

Ta có:

\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}\Leftrightarrow24+48y=18+72y\Leftrightarrow24=18+24y\)

\(\Rightarrow24y=6\Leftrightarrow y=\dfrac{1}{4}\)

Thay vào tìm được x

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

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

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

⇒\(9+3x=24\)

⇒\(3x=24-9=15\)

⇒\(x=15:3=5\)

Áp dụng t/c dãy tỉ số bằng nhau, ta có:

\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}=\dfrac{1+2y+1+4y+1+6y}{18+24+6x}=\dfrac{3+12y}{6\left(7+x\right)}=\dfrac{3\left(1+4y\right)}{2.3\left(7+x\right)}=\dfrac{1+4y}{2\left(7+x\right)}=\dfrac{1+4y}{24}\)=> 7 + x = 12

=> x = 5

Ta có :

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

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

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

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

\(\Leftrightarrow24y=6\)

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

Thay \(y=\dfrac{1}{4}\) ta có :

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

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

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

\(\Leftrightarrow6x=30\)

\(\Leftrightarrow x=5\)

Vạy ...

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

Từ \(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}\)=>\(24+48y=18+72y\)

=>24y=6=>\(y=\dfrac{1}{4}\)

Thay vào(1),ta có:

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

=>\(\dfrac{2}{24}=\dfrac{\dfrac{5}{2}}{6x}\)

=>12x=60=>x=5

Vậy...

TOÁN

\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}\Rightarrow\dfrac{2+4y}{36}=\dfrac{1+4y}{24}\)

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

\(\dfrac{1+6y}{6x}=\dfrac{2+4y}{36}=\dfrac{1+4y}{24}=\dfrac{\left(2+4y\right)-\left(1+4y\right)}{36-24}=\dfrac{1}{12}\)

\(\Rightarrow y=\dfrac{\dfrac{1}{12}.24-1}{4}=\dfrac{1}{4}=0,25\)

\(\Rightarrow\dfrac{1+6y}{6x}=\dfrac{1+1,5}{6x}\)

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

\(\Rightarrow6x=2,5:\dfrac{1}{12}=30\Leftrightarrow x=5\)

Vậy x = 5