Question

Biết x thỏa mãn \(\dfrac{x}{x^2+x+1}=\dfrac{1}{4}\). Tính \(\dfrac{x^5-3x^3-10x+12}{x^4+7x^2+15}\)

Answer 1

\(\dfrac{x}{x^2+x+1}=\dfrac{1}{4}\)

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

=>\(x^2-3x+1=0\)

\(\dfrac{x^5-3x^3-10x+12}{x^4+7x^2+15}\)

\(=\dfrac{x^5-3x^4+x^3+3x^4-9x^3+3x^2+5x^3-15x^2+5x+12x^2-36x+12+21x}{x^4+7x^2+15}\)

\(=\dfrac{x^3\left(x^2-3x+1\right)+3x^2\left(x^2-3x+1\right)+5x\left(x^2-3x+1\right)+12\left(x^2-3x+1\right)+21x}{x^4+7x^2+15}\)

\(=\dfrac{21x}{x^4-3x^3+x^2+3x^3-9x^2+3x+15x^2-45x+15+42x}\)

\(=\dfrac{21x}{x^2\left(x^2-3x+1\right)+3x\left(x^2-3x+1\right)+15\left(x^2-3x+1\right)+42x}\)

\(=\dfrac{21x}{42x}=\dfrac{1}{2}\)