a,[(x + 2)(x – 5)]^2016 = 0
⇒(x+2)(x-5)=0
⇒\(\left[ \begin{array}{l}x+2=0\\x-5=0\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=-2\\x=5\end{array} \right.\)
Vậy x ∈{-2;5}
.
b,(x – 5)^4 = 2^8
⇒(x – 5)^4=(2^2)^4
⇒x-5=2^2
⇒x-5=4
⇒x=9
Vậy x=9
.
c, xy + x – 2y = 3 (x, y ∈ Z)
⇒x(y+1)-2y-2=3-2
⇒x(y+1)-2(y+1)=1
⇒(x-2)(y+1)=1
TH1: $\left \{ {{x-2=-1} \atop {y+1=-1}} \right.$ ⇒$\left \{ {{x=1} \atop {y=-2}} \right.$
TH2: $\left \{ {{x-2=1} \atop {y+1=1}} \right.$ ⇒$\left \{ {{x=3} \atop {y=0}} \right.$
Vì (x, y ∈ Z)
⇒$\left \{ {{x=1} \atop {y=-2}} \right.$ hoặc $\left \{ {{x=3} \atop {y=0}} \right.$
