Question

(x-3)^3-(x-4)(x^2+4x+16)=49

Answer 1

Ta có: \(\left(x-3\right)^2-\left(x-4\right)\left(x^2+4x+16\right)=49\)

\(\Leftrightarrow x^3-9x^2+27x-27-x^3+64=49\)

\(\Leftrightarrow-9x^2+27x-12=0\)

\(\Leftrightarrow-3\left(x^2-9x+4\right)=0\)

\(\Delta=\left(-9\right)^2-4\cdot1\cdot4=81-16=65\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{9-\sqrt{65}}{2}\\x_2=\dfrac{9+\sqrt{65}}{2}\end{matrix}\right.\)