Tìm x,y biết rằng :
1+2y/18=1+4y/24=1+6y/6x
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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 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 \(\frac{1+2y}{18}\)=\(\frac{1+4y}{24}\)
\(\Rightarrow\)(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
\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}=\frac{\left(1+2y\right)+\left(1+6y\right)}{18+6x}\)(tính chất dãy tỉ số bằng nhau)
\(=\frac{2+8y}{2\left(9+3x\right)}\)
\(=\frac{2\left(1+4y\right)}{2\left(9+3x\right)}=\frac{1+4y}{9+3x}\)
\(\Rightarrow\frac{1+4y}{24}=\frac{1+4y}{9+3x}\Rightarrow9+3x=24\Rightarrow x=\frac{\left(24-9\right)}{3}=5\)
Vậy x = 5
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
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
Ý bạn là :\(1+\frac{2y}{18}=1+\frac{4y}{24}=1+\frac{6y}{6x}\)
hay \(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\)ạ ??
Lần sau ghi rõ :>
Ta có: \(\frac{1+2y}{18}=\frac{1+4y}{24}\)
=> \(\left(1+2y\right).24=\left(1+4y\right).18\)
=> \(24+48y=18+72y\)
=> \(24-18=72y-48y\)
=> \(24y=6\)
=> \(y=\frac{1}{4}\)
Với y = 1/4 => \(\frac{1+4\cdot\frac{1}{4}}{24}=\frac{1+6\cdot\frac{1}{4}}{6x}\)
=> \(\frac{1}{12}=\frac{\frac{5}{2}}{6x}\)
=> \(6x=\frac{5}{2}.12\)
=> \(6x=30\)
=> \(x=5\)