Tìm x biết : \(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)
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;z biết :
1) \(\dfrac{1+2y}{6}=\dfrac{3+4y}{5}=\dfrac{9+6y}{2x+1}\)
2) \(\dfrac{1+2y}{18}=\dfrac{1+4y}{28}=\dfrac{1+6y}{6x}\)
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
Tìm x; y
\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)
Ta có:\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}=\frac{2\left(1+2y\right)-\left(1+4y\right)}{2.18-24}=\frac{1+2y+1+4y-\left(1+6y\right)}{18+24-6x}\)
\(\Rightarrow\frac{1}{6}=\frac{1}{6x}\Rightarrow x=1\)
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
Tìm x biết : \(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)
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 ...
Tìm x , biết : \(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)
\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}\Rightarrow\dfrac{2+4y}{36}=\dfrac{1+4y}{24}\)
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
Tìm x,y biết:\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)
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
tìm x,y biết \(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)
\(\dfrac{1+2y}{18}=\dfrac{1+6y}{6x}\\ \Rightarrow\dfrac{1+2y}{18}=\dfrac{1+6y}{6x}=\dfrac{\left(1+2y\right)+\left(1+6y\right)}{18+6x}\\ =\dfrac{2+8y}{18+6x}=\dfrac{1+4y}{9+3x}\\ \Rightarrow9+3x=24\\ \Rightarrow3x=15\\ \Rightarrow x=5\)
Ta có: \(\dfrac{1+2y}{18}+\dfrac{1+4y}{24}\)
\(\Rightarrow24+48y=18+72y\)
\(\Rightarrow6=24y\Rightarrow y=\dfrac{1}{4}\)
\(\Rightarrow\dfrac{1+\dfrac{1}{2}}{18}+\dfrac{1+\dfrac{3}{2}}{6x}\)
\(\Rightarrow9x=45\Rightarrow x=5\)
tìm x biết:
\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)
\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)
\(\dfrac{1+2y}{18}=\dfrac{1+6y}{6x}=\dfrac{2+8y}{18+6x}=\dfrac{1+4y}{9+3x}\\ \Rightarrow9+3x=24\\ \Rightarrow3x=15\\ \Rightarrow x=5\)
Ta có : \(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)
Theo tính chất của dãy tỉ số bằng nhau :
\(\dfrac{1+2y+1+4y+1+6y}{18+24+6x}=\dfrac{3+12y}{42+6x}=\dfrac{3\left(1+4y\right)}{3\left(14+2y\right)}=\dfrac{1+4y}{14+2y}\)
Ta thấy :\(\dfrac{1+4y}{24}=\dfrac{1+4y}{14+2y}\)
\(\Leftrightarrow14+2y=24\Rightarrow2y=10\Leftrightarrow y=5\)
Xét \(\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)
\(\Rightarrow\dfrac{21}{24}=\dfrac{31}{6x}\)
\(\Rightarrow31.24=21.6x\Leftrightarrow744=126.x\)
\(\Rightarrow x=744:126=\dfrac{744}{126}\)
Tìm x,y biết :
\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)
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