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;z biết :
1) \(\dfrac{1+2y}{6}=\dfrac{3+4y}{5}=\dfrac{9+6y}{2x+1}\)
2) \(\dfrac{1+2y}{18}=\dfrac{1+4y}{28}=\dfrac{1+6y}{6x}\)
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, \(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{6x}\)
a,\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{+6x}\)
b, \(\dfrac{1+3y}{12}=\dfrac{1+5y}{5x}=\dfrac{1+7y}{4x}\)
c,\(\dfrac{x}{z+y+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y-2}=x+y+z\left(x,y,zkhac0\right)\)
d, \(\dfrac{3x}{8}=\dfrac{3y}{64}=\dfrac{3z}{216}va2x^2+2y^2-z^2=1\)
Tìm x , y , z biết :
\(\dfrac{1+2y}{18}=\dfrac{1+7y}{9x}=\dfrac{1+6y}{6x}\)
Tìm x , y , z :
\(\dfrac{1+2y}{18}=\dfrac{1+2y}{9x}=\dfrac{1+6y}{6x}\)
Tìm x, biết rằng : \(\dfrac{1+2x}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)
giúp mik vs
1. tìm x,y,z biết
a. 8x=5y,7y=12z và x+y+z= -318
b. \(\dfrac{5z-6y}{4}=\dfrac{6x-4z}{5}=\dfrac{4y-5x}{6}\)và 3x-2y+5z=96
c. 2009-|x-2009|=x
d. \(\left(2x-1\right)^{2008}\)+\(\left(y-\dfrac{2^{ }}{5}\right)^{2008}\)+|x+y-z|=0
Câu 1: Tìm x hoặc y:
\(\dfrac{1+2y}{18}\)=\(\dfrac{1+4y}{24}\)=\(\dfrac{1+6y}{6x}\)
Câu 2: Cho tỉ lệ thức \(\dfrac{a}{b}\)=\(\dfrac{b}{c}\)=\(\dfrac{c}{d}\)
Chứng minh rằng: \(\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\)
Bài x , y , z
a, y+x+1 /a = a + z +2 / y =x + y -3 /z =1/ x+ y +z
b, 1 + 2y/18 = 1 + 4y /24 = 1 + 6y / 6x
