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\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)(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
Ta có:\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}=\frac{2\left(1+2y\right)-\left(1+4y\right)}{2.18-24}=\frac{1+2y+1+4y-\left(1+6y\right)}{18+24-6x}\)
\(\Rightarrow\frac{1}{6}=\frac{1}{6x}\Rightarrow x=1\)
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
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
\(\dfrac{1+2y}{18}=\dfrac{1+6y}{6x}\\ \Rightarrow\dfrac{1+2y}{18}=\dfrac{1+6y}{6x}=\dfrac{\left(1+2y\right)+\left(1+6y\right)}{18+6x}\\ =\dfrac{2+8y}{18+6x}=\dfrac{1+4y}{9+3x}\\ \Rightarrow9+3x=24\\ \Rightarrow3x=15\\ \Rightarrow x=5\)
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...
\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}\)
khi và chỉ khi: \(\left(1+2y\right).24=\left(1+4y\right).18\) \(\Rightarrow24+48y=18+72y\)\(\Rightarrow72y-48y=24-18\)
\(\Rightarrow24y=6\) \(\Rightarrow y=\dfrac{1}{4}\) Lại có: \(\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\) \(\Rightarrow\dfrac{1+4\cdot\dfrac{1}{4}}{24}=\dfrac{1+6\cdot\dfrac{1}{4}}{6x}\) \(\Rightarrow\dfrac{1}{12}=\dfrac{\dfrac{5}{2}}{6x}\) \(\Rightarrow6x.1=12\cdot\dfrac{5}{2}\) \(\Rightarrow x=30:6=5\) Vậy x=5;y=\(\dfrac{1}{4}\)