Question

Tìm giá trị lớn nhất (hoặc nhỏ nhất) của các biểu thức sau:
a) A= x²-6x+11            b) B=2x²+10-1              c) C=5x-x²

Answer 1

A=11

B=9

C=0

Answer 2

a) \(A=x^2-6x+11\)

\(A=x^2-6x+9+2\)

\(A=\left(x-3\right)^2+2\)

Có: \(\left(x-3\right)^2\ge0\Rightarrow\left(x-3\right)^2+2\ge2\)

Dấu = xảy ra khi: \(\left(x-3\right)^2=0\Rightarrow x-3=0\Rightarrow x=3\)

Vậy: \(Min_A=2\) tại \(x=3\)

b) \(B=2x^2+10x-1\)

\(B=2x^2+10x+\frac{25}{2}-\frac{27}{2}\)

\(B=\left(\sqrt{2}x-\sqrt{\frac{25}{2}}\right)^2-\frac{27}{2}\)

Có: \(\left(\sqrt{2}x-\sqrt{\frac{25}{2}}\right)^2\ge0\Rightarrow\left(\sqrt{2}x-\sqrt{\frac{25}{2}}\right)^2-\frac{27}{2}\ge-\frac{27}{2}\)

Dấu = xảy ra khi: \(\left(\sqrt{2}x-\sqrt{\frac{25}{2}}\right)^2=0\Rightarrow\sqrt{2}x-\sqrt{\frac{25}{2}}=0\Rightarrow x=\frac{5}{2}\)

Vậy: \(Min_B=-\frac{27}{2}\) tại \(x=\frac{5}{2}\)

c) \(C=5x-x^2\) 

\(C=\frac{25}{4}-x^2+5x-\frac{25}{4}\)

\(C=\frac{25}{4}-\left(x-\frac{5}{2}\right)^2\)

Có: \(\left(x-\frac{5}{2}\right)^2\ge0\Rightarrow\frac{25}{4}-\left(x-\frac{5}{2}\right)^2\le\frac{25}{4}\)

Dấu = xảy ra khi: \(\left(x-\frac{5}{2}\right)^2=0\Rightarrow x-\frac{5}{2}=0\Rightarrow x=\frac{5}{2}\)

Vậy: \(Max_C=\frac{25}{4}\) tại \(x=\frac{5}{2}\)

Answer 3

a/ \(A=x^2-6x+11=\left(x^2-6x+9\right)+2=\left(x-3\right)^2+2\ge2\)

b/ \(B=2x^2+10-1=2x^2+9\ge9\)

c/ \(C=5x-x^2=\left(-x^2+\frac{2\times5x}{2}-\frac{25}{4}\right)+\frac{25}{4}\)

\(=\frac{25}{4}-\left(x-\frac{5}{2}\right)^2\le\frac{25}{4}\)

Answer 4

(14,78-a)/(2,87+a)=4/1

14,78+2,87=17,65

Tổng số phần bằng nhau là 4+1=5

Mỗi phần có giá trị bằng 17,65/5=3,53

=>2,87+a=3,53

=>a=0,66.

Answer 5

Bài làm:

a) \(A=x^2-6x+11=\left(x^2-6x+9\right)+2=\left(x-3\right)^2+2\ge2\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(x-3\right)^2=0\Rightarrow x=3\)

Vậy \(Min\left(A\right)=2\Leftrightarrow x=3\)

b) \(B=2x^2+10x-1=2\left(x^2+5x+\frac{25}{4}\right)-\frac{27}{2}=2\left(x+\frac{5}{2}\right)^2-\frac{27}{2}\ge-\frac{27}{2}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(2\left(x+\frac{5}{2}\right)^2=0\Rightarrow x=-\frac{5}{2}\)

Vậy \(Min\left(B\right)=-\frac{27}{2}\Leftrightarrow x=-\frac{5}{2}\)

c) \(C=5x-x^2=-\left(x^2-5x+\frac{25}{4}\right)+\frac{25}{4}=-\left(x-\frac{5}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(-\left(x-\frac{5}{2}\right)^2=0\Rightarrow x=\frac{5}{2}\)

Vậy \(Max\left(C\right)=\frac{25}{4}\Leftrightarrow x=\frac{5}{2}\)

Học tốt!!!!