Câu hỏi của Rosie

Question

tìm GTNN hoặc GTLN của :

A = a2 + b2 + c2 - ab -ac - bc +3

với a+b+c = 1

B = 5x - 6x2

Nguyễn Việt Lâm · Accepted Answer

$A=\frac{1}{2}\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)+3$

$=\frac{1}{2}\left(a^2-2ab+b^2\right)+\frac{1}{2}\left(b^2-2bc+c^2\right)+\frac{1}{2}\left(c^2-2ca+a^2\right)+3$

$=\frac{1}{2}\left(a-b\right)^2+\frac{1}{2}\left(b-c\right)^2+\frac{1}{2}\left(c-a\right)^2+3\ge3$

$A_{min}=3$ khi $a=b=c=1$

$B=-6\left(x^2-2.\frac{5}{12}x+\frac{25}{144}\right)+\frac{25}{24}$

$B=-6\left(x-\frac{5}{12}\right)^2+\frac{25}{24}\ge\frac{25}{24}$

$B_{min}=\frac{25}{24}$ khi $x=\frac{5}{12}$Câu trả lời: $A=\frac{1}{2}\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)+3$

$=\frac{1}{2}\left(a^2-2ab+b^2\right)+\frac{1}{2}\left(b^2-2bc+c^2\right)+\frac{1}{2}\left(c^2-2ca+a^2\right)+3$

$=\frac{1}{2}\left(a-b\right)^2+\frac{1}{2}\left(b-c\right)^2+\frac{1}{2}\left(c-a\right)^2+3\ge3$

$A_{min}=3$ khi $a=b=c=1$

$B=-6\left(x^2-2.\frac{5}{12}x+\frac{25}{144}\right)+\frac{25}{24}$

$B=-6\left(x-\frac{5}{12}\right)^2+\frac{25}{24}\ge\frac{25}{24}$

$B_{min}=\frac{25}{24}$ khi $x=\frac{5}{12}$

Violympic toán 8