Question

(2x-1)²⁰¹⁶ +(3y+6)²⁰¹⁴+(z-1)²⁰¹²=0

Vậy 4x+y-3z = ...........................

Answer 1

x=1/2,y=-2;z=1

Vậy 4x+y-3z=4.1/2+(-2)-3.1=-3

Answer 2

Ta có (2x-1)\(^{2016}\)+(3y+6)\(^{2014}\)+(z-1)\(^{2012}\)=0

\(\Leftrightarrow\)(2x-1)\(^{2016}\)=0 ; (3y+6)\(^{2014}\)=0 ; (z-1)\(^{2012}\)=0

Ta co :(2x-1)\(^{2016}\)=0\(\Rightarrow\)2x-1=0\(\Rightarrow\)2x=1\(\Rightarrow\)x=\(\frac{1}{2}\)

          (3y+6)\(^{2014}\)=0 \(\Rightarrow\)3y+6=0 \(\Rightarrow\)3y=-6 \(\Rightarrow\)y=-2

          (z-1)\(^{2012}\)=0 \(\Rightarrow\)z-1=0 \(\Rightarrow\)z=1

Vậy  4x+y-3z=4*\(\frac{1}{2}\)+(-2)-3*1=2-2-3=-3