Question

Tính giá trị của biểu thức

A= \(\dfrac{x^5-5x^3-4x+2}{x^4+x^2-14x-4}\) biết x² - 2x - 1 = 0

Answer 1

\(=\dfrac{x^5+2x^4-x^3-2x^4-4x^3+2x^2-2x^2-4x+2}{x^4-2x^3-x^2+2x^3-4x^2-2x+6x^2-12x-6+2}\)

\(=\dfrac{\left(x^2+2x-1\right)\left(x^3-2x^2-2\right)}{x^2\left(x^2-2x-1\right)+2x\left(x^2-2x-1\right)+6\left(x^2-2x-1\right)+2}\)

\(=\dfrac{\left(x^2+2x-1\right)\left(x^3-2x^2-2\right)}{2}\)

\(=\dfrac{x^5-2x^4-x^3+2x^4-4x^3-2x^2+2x^2-4x-2+4}{2}\)

\(=\dfrac{x^3\left(x^2-2x-1\right)+2x^2\left(x^2-2x-1\right)+2\left(x^2-2x-1\right)+4}{2}\)

=4/2=2