Tìm x,y biết \(\dfrac{1-2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)
Tìm x,y,z biết \(\dfrac{x}{3}=\dfrac{y}{4}\) ; \(\dfrac{y}{5}=\dfrac{z}{7}\) và x+y+z =9^2
Tìm n để A= 12n^2 - 5n -25 là số nguyên tố
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 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 :
\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)
Áp dụng tính chất của dãy tỉ số bằng nhau , ta có :
\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}=\dfrac{1+2y+1+6y}{18+6x}=\dfrac{2.\left(1+4y\right)}{2.\left(9+3x\right)}=\dfrac{1+4y}{9+3x}\)
⇒\(\dfrac{1+4y}{9+3x}=\dfrac{1+4y}{24}\)
⇒\(9+3x=24\)
⇒\(3x=24-9=15\)
⇒\(x=15:3=5\)
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 , y ,z :
a, \(\dfrac{x+z+1}{x}=\dfrac{z+x+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
b, 10x = 6y và \(2x^2-y^2=-28\)
c, \(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)
d, \(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{6x}\)
Ta có : 2x+1 /5 = 3y-2/7 = 2x+3y -1 /6x
=> 2x+1+3y-2 / 5+7 = 2x+3y-1 /6x
=> 2x+3y-1 / 12 = 2x+3y-1 / 6x
=> 12 = 6x => x =2
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 ; y ;z :
a. \(\dfrac{x}{3}=\dfrac{y}{4}\) ; 7y = 5z và 2x+3y -z = 186
b. \(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)
a, Ta có: \(7y=5z\Leftrightarrow\dfrac{y}{5}=\dfrac{z}{7}\)
Ta lại có: \(\dfrac{x}{3}=\dfrac{y}{4}\Leftrightarrow\dfrac{x}{15}=\dfrac{y}{20}\left(1\right)\)
\(\dfrac{y}{5}=\dfrac{z}{7}\Leftrightarrow\dfrac{y}{20}=\dfrac{z}{28}\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\Leftrightarrow\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}\) và \(2x+3y-z=186\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{186}{62}=3\)
+) \(\dfrac{2x}{30}=3\Rightarrow2x=3.30=90\Rightarrow x=90:2=45\)
+) \(\dfrac{3y}{60}=3\Rightarrow3y=3.60=180\Rightarrow y=180:3=60\)
+) \(\dfrac{z}{28}=3\Rightarrow z=3.28=84\)
Vậy ...
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