Question

tìm x nguyên:    \(x^2-3x+5=3\)

Answer 1

x²-3x+5=3

x²-3x+2=0

x²-2x-x+2=0

x(x-2)-(x-2)=0

(x-2)(x-1)=0

=>x-2=0 hoặc x-1=0

x=2 hoặc x=1

Answer 2

\(x^2-3x+5=3\)

\(x^2-2x-x+5-3=0\)

\(x^2-x-2x+2=0\)

\(x\left(x-1\right)-2\left(x-1\right)=0\)

\(\left(x-1\right).\left(x-2\right)=0\)

\(\Rightarrow x-1=0\)hoặc \(x-2=0\)

\(x=0+1\)    hoặc     \(x=0+2\)

\(x=1\) hoặc \(x=2\)

Vậy \(x=1\) hoặc \(x=2\)