Question

giải phương trình sau:

\(3\left(x^2-6x+10\right)^2+2\left(x^2-6x\right)-65=0\)

Answer 1

\(3\left(x^2-6x+10\right)^2+2\left(x^2-6x\right)-65=0\\ \Leftrightarrow3\left(\left(x-3\right)^2+1\right)^2+2\left(x^2-6x+9\right)-83=0\\ \Leftrightarrow3\left(\left(x-3\right)^4+2.\left(x-3\right)^2+1\right)+2\left(x-3\right)^2-83=0\\ \Leftrightarrow3\left(x-3\right)^4+6\left(x-3\right)^2+3+2\left(x-3\right)^2-83=0\\ \Leftrightarrow3\left(x-3\right)^4+8\left(x-3\right)^2-80=0\\ \Leftrightarrow\left(x-3\right)^2\left(3.\left(x-3\right)^2+8\right)=80\\ \Leftrightarrow\left(x-3\right)^2.\left(3x^2-18x+27+8\right)=80\)

Phá ra rồi giải tiếp !!