Question

tìm x,y,z biết :

6x=4y=3z và x+y+z=18

 

Answer 1

Ta có : \(6x=4y=3z=\frac{x}{\frac{1}{6}}+\frac{y}{\frac{1}{4}}+\frac{z}{\frac{1}{3}}=\frac{x+y+z}{\frac{1}{6}+\frac{1}{4}+\frac{1}{3}}=\frac{18}{\frac{3}{4}}=24\)

Nên : 6x = 24 = > x = 4

        4y = 24 => y = 6

         3z = 24 => z = 8

Vậy x = 4 ; y = 6 ; z = 8

Answer 2

Ta có: 6x=4y=3z

6x=4y

=>\(\frac{x}{4}=\frac{y}{6}\)

4y=3z

=>\(\frac{y}{3}=\frac{z}{4}\)

=>\(\frac{y}{6}=\frac{z}{8}\)

=>\(Đặt\frac{x}{4}=\frac{y}{6}=\frac{z}{8}=k\)

x=4k;y=6k;z=8k

=>x+y+z=4k+6k+8k

=>x+y+z=18k=18

=>k=1

=>x=4k=4.1=4 ; y=6k=6.1=6 ; z=8k=8.1=8

vậy....

nha