Question

cho hàm số \(y=\dfrac{3x^2+2x+1}{x-2}\) có đạo hàm là biểu thức có dạng \(\dfrac{ax^2+bx+c}{\left(2x-4\right)^2}\). tinh \(a^2-b^2+c^2\)

Answer 1

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

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

\(y'=\dfrac{6x^2-10x-4-3x^2-2x-1}{\left(x-2\right)^2}=\dfrac{3x^2-12x-5}{\left(x-2\right)^2}=\dfrac{12x^2-48x-20}{\left(2x-4\right)^2}\)

\(\Rightarrow a^2-b^2+c^2=12^2-48^2+20^2=...\)