Question

bài 1 tìm giá trị nhỏ nhất

A= 2x²+10x -1

Answer 1

A=2x²+10x-1

A=2(x²+5x-\(\frac{1}{2}\))

A=2[x²+2x*\(\frac{5}{2}\)+(\(\frac{5}{2}\))²-(\(\frac{5}{2}\))²-\(\frac{1}{2}\)]

A=2[(x+\(\frac{5}{2}\))²-\(\frac{27}{4}\)]

A=2(x+\(\frac{5}{2}\))²-\(\frac{27}{2}\)

Ta có: 2(x+\(\frac{5}{2}\))²≥0

⇒ 2(x+\(\frac{5}{2}\))²-\(\frac{27}{2}\)≥\(\frac{-27}{2}\)

⇒ A_min=\(\frac{-27}{2}\) khi x+\(\frac{5}{2}\)=0⇒x=\(\frac{-5}{2}\).

Hơi dài nhưng đầy đủ nha!!!!!