Question

Cho \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)và \(2x+3y-z=50\)

Tính x+y+z ?

Answer 1

\(\frac{x-1}{2}\)=\(\frac{2\left(x-1\right)}{2}\)=\(\frac{2x-2}{2}\)

\(\frac{y-2}{3}\)=\(\frac{3\left(y-2\right)}{3}\)=\(\frac{3y-6}{3}\)

=> \(\frac{2x-2}{4}\)=\(\frac{3y-6}{9}\)=\(\frac{z-3}{4}\)=\(\frac{2x-2+3y-6-\left(z-3\right)}{4+9-4}\)

= \(\frac{2x+3y-z+3-2-6}{9}\)=\(\frac{50-5}{9}\)=\(\frac{45}{9}\)=\(5\)

x-1 = 5.2 = 10

=> x=11

y-2 = 5.3 = 15

=> y=17

z-3 = 5.4 = 20

=> z=23

 Vậy z+y+z= 11+17+23 = 51

Tik cho mk nha................cảm ơn rất nhiều

 

Answer 2

x=11 ; y = 17 ; z = 23 => x+y+z = 11+17+23 = 51