Question

giải pt

\(\left(x^2-3x+1\right)\left(x^2+3x+2\right)\left(x^2-9x+20\right)=-30\)

Answer 1

\(\Leftrightarrow\left(x^2-3x+1\right)\left(x+1\right)\left(x+2\right)\left(x-4\right)\left(x-5\right)=-30\)

\(\Leftrightarrow\left(x^2-3x+1\right)\left(x^2-3x-4\right)\left(x^2-3x-5\right)=-30\)

Đặt x^2-3x=a

=>(a+1)(a-4)(a-5)=-30

=>\(\left(a^2-3a-4\right)\left(a-5\right)=-30\)

=>\(a^3-5a^2-3a^2+15a-4a+20+30=0\)

=>a^3-8a^2+11a+50=0

=>a=-1,77

=>x^2-3x=-1,77

=>x^2-3x+1,77=0

hay \(\left[{}\begin{matrix}x=\dfrac{15+4\sqrt{3}}{10}\\x=\dfrac{15-4\sqrt{3}}{10}\end{matrix}\right.\)