Question

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, \(\dfrac{3x}{8}=\dfrac{3y}{64}=\dfrac{3z}{216}va2x^2+2y^2-z^2=1\)

Answer 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

Answer 2

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

Answer 3

Mình sửa lại câu a

1+2y/18 = 1+6y / 6x = 1+2y+1+6y / 18 + 6x = 2 +8y /18+6x

= 2 (1+4y) / 2 (9 +3x) = 1+4y / 9 +3x

Ta lại có: 1+4y/24 = 1+4y/ 9 +3x

=> 24 = 9 +3x => 15= 3x => x =5

Answer 4

x/x+y+1 = y/ x+z+1 = z/x+y-2 = x+y+z/ x+y+1+x+z+1+x+y-2

=x+y+z/ 2(x+y+z) = 1/2

Nếu x+y+z = 0 => x=0; y=0; z=0

Nếu x+y+z khác 0 => x+y+z=1/2

=> x+y = 1/2-z => y+z= 1/2 -x => x+z = 1/2 - y

Ta có: x/ 1/2 -x +1 = 1/2

=> 2x = 1/2-x+1

=>3x= 1/2+1

=> 3x=3/2 => x=1/2

y/ 1/2-y+1 => 2y=1/2-y+1=>3y=1/2+1=>3y=3/2=>y=1/2

z/ 1/2-z-2 => 2z = 1/2-z-2 => 3z= 1/2-2 => 3z= -3/2=> z= -1/2