Question

Cho \(x^2-x=3\) Tính giá trị của biểu thức:.

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

Answer 1

Ta có  : \(M=x^4-2x^3+3x^2-2x+2\)

\(=\left(x^4-2x^3+x^2\right)+\left(2x^2-2x\right)+2\)

\(=\left(x^2-x\right)^2+2\left(x^2-x\right)+2\)

\(=3^2+2.3+2=9+6+2=17\)

Answer 2

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

     \(=\left(x^4-x^3\right)-\left(x^3-x^2\right)+\left(2x^2-2x\right)+2\)

     \(=x^2\left(x^2-x\right)-x\left(x^2-x\right)+2\left(x^2-x\right)+2\)

     \(=\left(x^2-x\right)\left(x^2-x+2\right)\)

\(M=\left(x^2-x\right)\left(x^2-x+2\right)\)

\(\rightarrow M=3\left(3+2\right)=15\)