Question

Ba số x,y,z thỏa mãn x-y=8 ;y-z=10;x+z=12 . khi do x+Y+Z

Answer 1

Ta có:

\(x-y=8\)

\(y-z=10\)

\(x+z=12\)

\(\Rightarrow\left(x-y\right)+\left(y-z\right)+\left(x+z\right)=8+10+12\)

\(\Rightarrow x-y+y-z+x+z=30\)

\(\Rightarrow x+x-y+y-z+z=30\)

\(\Rightarrow2x=30\)

\(\Rightarrow x=30\div2\)

\(\Rightarrow x=15\)

Thay \(x=15\) vào \(x-y=8\) ta được:

\(15-y=8\)

\(\Rightarrow y=15-8\)

\(\Rightarrow y=7\)

Thay \(y=7\) vào \(y-z=10\) ta được:

\(7-z=10\)

\(\Rightarrow z=7-10\)

\(\Rightarrow z=-3\)

\(\Rightarrow x+y+z=15+7+\left(-3\right)=19\)

Vậy tổng \(x+y+z=19\)

Answer 2

\(\left\{\begin{matrix}x-y=8\left(1\right)\\y-z=10\\x+z=12\left(3\right)\end{matrix}\right.\left(2\right)\)

(2)+(3)=> x+y=22 (4)

(3)-(1)=> y+z=4 (5)

(3)+(4)+(5)=2(x+y+z)=12+22+4

=> (x+y+z)=6+11+2=19