tim x,y biet
1+2y/18=1+4y/24=1+6y/6x
Tìm x , y , x biết 1 + 2y / 18 = 1 + 4y / 24 = 1 + 6y / 6x
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+2y181+2y18=1+4y241+4y24
⇒⇒(1+2y)24=(1+4y)18
⇒⇒24+48y=18+72y
⇒⇒24-18=72y-48y
⇒⇒6=24y
⇒⇒y=624624
⇒⇒y=1414
Thay y=1414 vào đề ta có:
1 + 21414 / 18 = 1 + 41414 / 24 = 1 + 61414 / 6x
=>112112=526x526x
=>12.5252=6x
=>30=6x
=>x=5
Vậy x=5;y=14
Tìm x; y ;z biết 1+2y/18=1+4y/24=1+6y/6x
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
Tìm x , y biết :
1 + 2y / 18 = 1 + 4y / 24 = 1 + 6y / 6x
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
Tìm x, y biết rằng: \(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)
\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)(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
Tìm x,y biết rằng :
1+2y/18=1+4y/24=1+6y/6x
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 => \(\frac{1+4\cdot\frac{1}{4}}{24}=\frac{1+6\cdot\frac{1}{4}}{6x}\)
=> \(\frac{1}{12}=\frac{\frac{5}{2}}{6x}\)
=> \(6x=\frac{5}{2}.12\)
=> \(6x=30\)
=> \(x=5\)
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
tìm x,y,z: biết (1+2y)/18=(1+4y)/24=(1+6y)/6x=(1+8y)/(27z)
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
Tìm x biết: (1+2y)/18=(1+4y)/24=(1+6y)/6x
(1+2y)/18=(1+6y)/6x=(1+2y+1+6y)/18+6x=(2+8y)/18+6x=(1+4y)/9+3x
=>(1+4y)/9+3x=(1+4y)/24 =>9+3x=24 =>3x=15 =>x=5
(1+2y)/18=(1+6y)/6x=(1+2y+1+6y)/18+6x=(2+8y)/18+6x=(1+4y)/9+3x
=>(1+4y)/9+3x=(1+4y)/24 =>9+3x=24 =>3x=15 =>x=5
Nhé bạn !
\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)
Ta có :
\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}\)
\(\Leftrightarrow24\left(1+2y\right)=18\left(1+4y\right)\)
\(\Leftrightarrow24+48y=18+72y\)
\(\Leftrightarrow72y-48y=24-18\)
\(\Leftrightarrow24y=6\)
\(\Leftrightarrow y=\dfrac{6}{24}=\dfrac{1}{4}\)
mà \(\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)
\(\Leftrightarrow6x\left(1+4y\right)=24\left(1+6y\right)\)
\(\Leftrightarrow6x\left(1+4.\dfrac{1}{4}\right)=24\left(1+6.\dfrac{1}{4}\right)\left(thay.y=\dfrac{1}{4}.vào\right)\)
\(\Leftrightarrow6x.2=24\left(1+\dfrac{3}{2}\right)\)
\(\Leftrightarrow12x=24.\dfrac{5}{2}\)
\(\Leftrightarrow12x=60\)
\(\Leftrightarrow x=5\)
Vậy \(\left\{{}\begin{matrix}x=5\\y=\dfrac{1}{4}\end{matrix}\right.\)
Tìm x,y biết
a,x+2y/18=1+4y/24=1+6y/6x
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
Tìm x biết (1+2y)/18=(1+4y)/24=(1+6y)/6x
xét:\(\frac{1+2y}{18}=\frac{1+6y}{6x}=\frac{1+2y+1+6y}{18+6x}=\frac{2+8y}{18+6x}=\frac{2\left(1+4y\right)}{2\left(9+3x\right)}=\frac{1+4y}{9+3x}\) (t/c dãy tỉ số=nhau)
=>\(\frac{1+4y}{24}=\frac{1+4y}{9+3x}\Rightarrow24=9+3x\Rightarrow x=5\)
vậy x=5 thỏa mãn
Ta có: 1 + 2y/18 = 1 + 4y/24 = 1 + 6y/6x = 1 + 2y + 1 + 6y/18 + 6x = 2 + 8y/18 + 6x = 2(1 + 4y)/18 + 6x
\ 18 + 6x = 2 . 24 = 48
\ 6x = 30
\ x = 5
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)
tìm x biết rằng 1+2y/18=1+4y/24=1+6y/6x
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
cho 3 tỉ số = nhau là : a/b+c; b/c+a; c/a+b
tìm gt của mỗi tỉ số đó
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