Question

R=799\(x^2\)-3\(x^4\)+403x+1198\(x^3\)-1203 với x=400

Answer 1

Nếu \(x=400\Rightarrow\left\{{}\begin{matrix}799=2x-1\\403=x+3\\1198=3x-2\\1203=3\left(x+1\right)\end{matrix}\right.\)( * )

Thay ( * ) vào R , ta được :

\(R=\left(2x-1\right)x^2-3x^4+\left(x+3\right)x+\left(3x-2\right)x^3-3\left(x+1\right)\)

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

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

\(=-3\)

Vậy \(R=-3\) tại \(x=400\)