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, \(\dfrac{3x}{8}=\dfrac{3y}{64}=\dfrac{3z}{216}va2x^2+2y^2-z^2=1\)
a, 1+2y / 18 = 1+4y / 24 = 1+6y / 6x
Ta có : 1+2y / 18 = 1+6y / 6x = 1+2y + 1+6y / 18 + 6y
= 2+ 8y / 18+6y = 2 (1+4y) / 2( 9 +3y) = 1+4y/9+3y
Ta lại có : 1 + 4y/24 = 1+4y / 9+3y
=> 24=9+3y => 15=3y => y=5
Vậy y=5
Nhớ like
b, 1+3y/12 = 1+5y/5x = 1+7y/4x
Ta có : 1+3y/12 = 1+7y/4x = 1+3y+1+7y / 12 +4x
= 2 + 10y / 12 +4x = 2 (1+5y) / 2 (6+2x) = 1+5y / 6+2x
Ta lại có: 1+5y / 5x = 1+5y / 6+2x
=> 5x = 6+2x => 3x = 6 => x=2
Vậy x =2