Question

Tìm x

x⁸-x⁵+x²-x+1=0

Answer 1

\(VT=x^8-x^5+x^2-x+1\)

\(=x^8-2x^4\cdot\dfrac{x}{2}+\dfrac{x^2}{4}+\dfrac{x^2}{2}+x^4-x+1\)

\(=\left(x^4-\dfrac{x}{2}\right)^2+\left(\dfrac{x}{2}-1\right)^2+\dfrac{x^2}{2}>0\)

Hay pt vô nghiệm .-.

Answer 2

x⁸-x⁵+x²-x+1

=(x⁴-\(\dfrac{x^2}{2}\))²+\(\dfrac{3}{4}\)x²-x+1

=(x⁴-\(\dfrac{x^2}{2}\))+\(\dfrac{3}{4}\)(x²-\(\dfrac{4}{3}x\)+\(\dfrac{4}{3}\))

=(x⁴-\(\dfrac{x^2}{2}\))+\(\dfrac{3}{4}\)(x-2/3)²+\(\dfrac{2}{3}\)>0

=> Vô nghiệm