Question

Tìm giá trị lớn nhất:

B= (x²+10x+20):(x²+6x+9)

C= (3x²+9x+17):(3x²+9x+7)

Answer 1

\(B=\frac{x^2+10x+20}{x^2+6x+9}=\frac{(x^2+6x+9)+4(x+3)-1}{x^2+6x+9}\)

\(=1+\frac{4(x+3)}{x^2+6x+9}-\frac{1}{x^2+6x+9}=1+\frac{4(x+3)}{(x+3)^2}-\frac{1}{(x+3)^2}\)

\(=1+\frac{4}{(x+3)}-\frac{1}{(x+3)^2}\)

Đặt \(\frac{1}{x+3}=a\Rightarrow B=1+4a-a^2=5-(a^2-4a+4)\)

\(=5-(a-2)^2\leq 5\)

Vậy \(B_{\max}=5\Leftrightarrow a=2\Leftrightarrow x=-\frac{5}{2}\)

Answer 2

\(C=\frac{3x^2+9x+17}{3x^2+9x+7}=\frac{3x^2+9x+7+10}{3x^2+9x+7}=1+\frac{10}{3x^2+9x+7}\)

Có: \(3x^2+9x+7=3(x^2+3x+\frac{9}{4})+\frac{1}{4}=3(x+\frac{3}{2})^2+\frac{1}{4}\geq \frac{1}{4}\)

\(\Rightarrow \frac{10}{3x^2+9x+7}\leq \frac{10}{\frac{1}{4}}=40\)

\(\Rightarrow C\leq 41\)

Vậy \(C_{\max}=41\Leftrightarrow x=\frac{-3}{2}\)