Question

Answer 1

a: \(\left(x-1\right)^2+1=-2x^2+8\)

=>\(x^2-2x+1+1+2x^2-8=0\)

=>\(3x^2-2x-6=0\)

\(\text{Δ}=\left(-2\right)^2-4\cdot3\cdot\left(-6\right)=4+72=76>0\)

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

\(\left\{{}\begin{matrix}x_1=\dfrac{2-\sqrt{76}}{2\cdot3}=\dfrac{2-2\sqrt{19}}{6}=\dfrac{1-\sqrt{19}}{3}\\x_2=\dfrac{2+\sqrt{76}}{2\cdot3}=\dfrac{1+\sqrt{19}}{3}\end{matrix}\right.\)

b: \(x^3\left(x+1\right)-7=x^2\left(x+7\right)+1\)

=>\(x^4+x^3-7-x^3-7x^2-1=0\)

=>\(x^4-7x^2-8=0\)

=>\(\left(x^2-8\right)\left(x^2+1\right)=0\)

=>\(x^2-8=0\)

=>\(x^2=8\)

=>\(x=\pm2\sqrt{2}\)