Question

[bin nguyễn]

x mũ 4 cộng x mũ 3 cộng x cộng 1 = 4x mũ 2

Answer 1

[Trần Tuấn Hoàng]

\(x^4+x^3+x+1=4x^2\)

⇔\(x^4+x^3-4x^2+x+1=0\)

⇔\(\left(x^3-2x^2+x\right)+\left(x^4-2x^2+1\right)=0\)

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

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

⇔\(\left(x-1\right)^2\left[x\left(x+1\right)^2\right]=0\)

⇔\(\left(x-1\right)^2\left(x^2+3x+1\right)=0\)

⇔\(\left(x-1\right)^2=0\) hay \(x^2+3x+1=0\)

⇔\(x=1\) hay \(x^2+2.\dfrac{3}{2}x+\dfrac{9}{4}-\dfrac{5}{4}=0\)

⇔\(x=1\) hay \(\left(x+\dfrac{3}{2}\right)^2-\left(\dfrac{\sqrt{5}}{2}\right)^2=0\).

⇔\(x=1\) hay \(\left(x+\dfrac{3}{2}+\dfrac{\sqrt{5}}{2}\right)\left(x+\dfrac{3}{2}-\dfrac{\sqrt{5}}{2}\right)=0\)

⇔\(x=1\) hay \(x=-\dfrac{3+\sqrt{5}}{2}\) hay \(x=-\dfrac{3-\sqrt{5}}{2}\).

-Vậy \(S=\left\{1;-\dfrac{3+\sqrt{5}}{2};-\dfrac{3-\sqrt{5}}{2}\right\}\).