Question

Tìm GTLN của biểu thức:

a) A = 5-3. (2x - 1)²

b) B= \(\dfrac{1}{2.\left(x-1\right)^2+3}\)

c) C = \(\dfrac{x^2+8}{x^2+}\)

Answer 1

c)C=\(\dfrac{x^2+8}{x^2+2}=\dfrac{\left(x^2+2\right)+6}{x^2+2}=1+\dfrac{6}{x^2+2}\)

Để C đạt GTLN thì \(\dfrac{6}{x^2+2}\) đạt GTNN

\(x^2\ge0\Rightarrow x^2+2\ge2\)

Max C=4 khi x=0

Answer 2

a)A= 5-3.\(\left(2x-1\right)^2\)

\(\left(2x-1\right)^2\)\(\ge0\) nên 3.\(\left(2x-1\right)^2\)\(\ge0\)

Max A=5 khi x=\(\dfrac{1}{2}\)

b) Để B=\(\dfrac{1}{2.\left(x-1\right)^2+3}\)đạt GTLN thì \(2.\left(x-1\right)^2+3\) đạt GTNN

\(\left(x-1\right)^2\ge0\Rightarrow2.\left(x-1\right)^2\ge0\Rightarrow2.\left(x-1\right)^2+3\ge3\)

Max B=\(\dfrac{1}{3}\)khi x=1

câu c thiếu đề phải ko bạn