Question

Tìm x:  \(\left(x-2\right)^3+3\left(2x-3\right)^2=\left(3+x\right)^3-5\left(1+3x\right)^2-\left(x-1\right)\left(3-x\right)\)

Answer 1

\(\Leftrightarrow x^3-6x^2+12x-8+3\left(4x^2-12x+9\right)=x^3+9x^2+27x+27-5\left(9x^2+6x+1\right)+\left(x-1\right)\left(x-3\right)\)

\(\Leftrightarrow-6x^2+12x-8+12x^2-36x+27=9x^2+27x+27-45x^2-30x-5+\left(x-1\right)\left(x-3\right)\)

\(\Leftrightarrow6x^2-24x+19=-36x^2-3x+22+\left(x-1\right)\left(x-3\right)\)

\(\Leftrightarrow42x^2-21x-3-x^2+4x-3=0\)

\(\Leftrightarrow41x^2-17x-6=0\)

\(\Delta=\left(-17\right)^2-4\cdot41\cdot\left(-6\right)=1273\)

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

\(\left\{{}\begin{matrix}x_1=\dfrac{17-\sqrt{1273}}{82}\\x_2=\dfrac{17+\sqrt{1273}}{82}\end{matrix}\right.\)