Question

Tìm x

(X-1)^3 + 3(x-3)^2 -(x+2)(x^2-2x+4) = (x+2)^3 -(x-3)(x^2+9) - 6x^2 + 5

Answer 1

\(\Leftrightarrow x^3-3x^2+3x-1-x^3-8+3\left(x^2-6x+9\right)=x^3+6x^2+12x+8-x^3-9x+3x^2+27-6x^2+5\)

\(\Leftrightarrow-3x^2+3x-9+3\left(x^2-6x+9\right)=3x^2+3x+40\)

\(\Leftrightarrow-3x^2+3x-9+3x^2-18x+27-3x^2-3x-40=0\)

\(\Leftrightarrow-3x^2-18x-22=0\)

\(\Leftrightarrow3x^2+18x+22=0\)

\(\Delta=18^2-4\cdot3\cdot22=60\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-18-2\sqrt{15}}{6}=\dfrac{-9-\sqrt{15}}{3}\\x_2=\dfrac{-18+2\sqrt{15}}{6}=\dfrac{-9+\sqrt{15}}{3}\end{matrix}\right.\)