Question

Biết \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{ax^2+4x+8}-\sqrt[3]{81x^2+63x-19}}{x^3-x^2-x+1}=\dfrac{b}{c}\). Tính a+b+c

Answer 1

\(\sqrt{a+12}-\sqrt[3]{81+63-19}=0\Rightarrow a=13\)

Khi đó

\(\dfrac{\sqrt{13x^2+4x+8}-\sqrt[3]{81x^2+63x-19}}{\left(x-1\right)^2\left(x+1\right)}\)

\(=\dfrac{\sqrt[]{13x^2+4x+8}-\left(3x+2\right)+\left(3x+2-\sqrt[3]{81x^2+83x-19}\right)}{\left(x-1\right)^2\left(x+1\right)}\)

\(=\dfrac{\dfrac{4\left(x-1\right)^2}{\sqrt[]{13x^2+4x+8}+\left(3x+2\right)}+\dfrac{27\left(x-1\right)^2\left(x+1\right)}{\left(3x+2\right)^2+\left(3x+2\right)\sqrt[3]{81x^2+63x-19}+\sqrt[3]{\left(81x^2+63x-19\right)^2}}}{\left(x-1\right)^2\left(x+1\right)}\)