Question

Tìm x, y, z:

a, 2x/3 = 3y/ 4 = 4z/5 và x+2y+4z = 220

Answer 1

Đặt 2x/3+3y/4=4z/5=k   (k khác 0 )

<=>x= 3/2.k ;  y=4/3.k    ; z=5/4.k

mà x+2y+4z = 220

suy ra 3/2.k+2.4/3.k+4.5/4k = 220

<=>k = 24

Answer 2

\(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}=k\)

=>\(\begin{cases}2x=3k\\3y=4k\\4z=5k\end{cases}\)

=>\(\begin{cases}x=\frac{3k}{2}\\y=\frac{4k}{3}\\z=\frac{5k}{4}\end{cases}\)

ta có:

x+2y+4z=220

=> \(\frac{3k}{2}+2\left(\frac{4k}{3}\right)+4\left(\frac{5k}{4}\right)=220\)

=> \(\frac{3k}{2}+\frac{8k}{3}+5k=220\)

=> k(\(\frac{3}{2}+\frac{8}{3}+5\))=220

=> 55/6k=220

=> k=220.6/55=24

vậy

x=24.3/2=36

y=24.4/3=32

z=24.5/4=30

Answer 3

Answer 4