Question

chứng minh vô nghiệm : 1+x+x^2

Answer 1

x² + x + 1 = x² + \(\frac{1}{2}\). x+ \(\frac{1}{2}\).x + \(\frac{1}{4}\)+ \(\frac{3}{4}\) = (x² + \(\frac{1}{2}\). x) +( \(\frac{1}{2}\).x + \(\frac{1}{4}\)) + \(\frac{3}{4}\) = x.(x + \(\frac{1}{2}\) ) + \(\frac{1}{2}\).(x + \(\frac{1}{2}\)) + \(\frac{3}{4}\)

= (x + \(\frac{1}{2}\) ). (x + \(\frac{1}{2}\) ) + \(\frac{3}{4}\) = (x + \(\frac{1}{2}\))²  + \(\frac{3}{4}\) \(\ge\) 0 + \(\frac{3}{4}\)= \(\frac{3}{4}\) với mọi x

=> x² + x + 1 = 0 không có nghiệm