Question

tim GTNN A=x^4-x^2+7

Answer 1

A=x^4-x^2+7

=x⁴-2x²\(\frac{1}{2}\)+\(\frac{1}{4}\)+\(\frac{27}{4}\)

=(x²-\(\frac{1}{2}\))²+\(\frac{27}{4}\)

Vì (x²-\(\frac{1}{2}\))²\(\ge\)0 nên (x²-\(\frac{1}{2}\))²+\(\frac{27}{4}\)\(\ge\frac{27}{4}\)

Dấu = xảy ra khi x²-\(\frac{1}{2}\)=0

                       <=>x²=\(\frac{1}{2}\)

                      <=>x=\(\sqrt{\frac{1}{2}}\)hoặc x=\(-\sqrt{\frac{1}{2}}\)

Vậy GTNN của A là \(\frac{27}{4}\)tại x=\(\sqrt{\frac{1}{2}}\);\(-\sqrt{\frac{1}{2}}\)