Question

Tìm x (x²+1)(x-2)+2x=4 Mong mn giúp mình. Cảm ơn ạ

Answer 1

\(\left(x^2+1\right)\left(x-2\right)+2x=4\Leftrightarrow x^3-2x^2+x-2+2x-4=0\Leftrightarrow x^3-2x^2+3x-6=0\Leftrightarrow\left(x-2\right)\left(x^2+3\right)=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)(do \(x^2+3\ge3>0\))

Answer 2

Answer 3

\(x(x^2+1)(x-2)+2x=4 \)

\((x^3+x)(x-2)+2x-4=0\)

\(x^4-2x^3+x^2-2x+2x-4=0\)

\(x^4-2x^3+x^2-4=0\)

\((x^4-2x^3)+(x^2-4)=0\)

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

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

th1:\(x^3+x+2=0 \)

              x=-1

th2 x-2=0

         x=2

ptrình cs tập nghiệm S={-1,0}