Question

Cho hai biểu thức \(P=\dfrac{79x^2+1990x+142431}{x^3-5x^2+2006x-10030}\) ; Q=\(\dfrac{ax+b}{x^2+2006}+\dfrac{c}{x-5}\)

1) Xác định a,b,c để P=Q với mọi x≠5

2) Tính giá trị của P khi x = \(\dfrac{2005}{2006}\)

Answer 1

1: \(Q=\dfrac{\left(ax+b\right)\left(x-5\right)+c\left(x^2+2006\right)}{\left(x-5\right)\left(x^2+2006\right)}\)

\(=\dfrac{ax^2-5a+bx-5b+cx^2+2006c}{\left(x-5\right)\left(x^2+2006\right)}\)

\(=\dfrac{x^2\left(a+c\right)+bx-5a-5b+2006c}{\left(x-5\right)\left(x^2+2006\right)}\)

=>a+c=79; b=1990; -5a-5b+2006c=142431

=>a+c=79; -5a+2006c=152381; b=1990

=>a=6093/2011; c=75,97

2: Khi x=2005/2006 thì \(P=\dfrac{79\cdot\left(\dfrac{2005}{2006}\right)^2+1990\cdot\dfrac{2005}{2006}+142431}{\left(\dfrac{2005}{2006}-5\right)\left(\dfrac{2005}{2006}^2+1\right)}\)

=-18069,12068