Toán Tìm GTLN của A= 9/ x^2-x+5 B= 27-12x/ x^2+9 30/06/2021 By Samantha Tìm GTLN của A= 9/ x^2-x+5 B= 27-12x/ x^2+9
Đáp án: $max_A= \dfrac{36}{19}\Leftrightarrow x=\dfrac{1}{2}\\ max_B= 4\Leftrightarrow x=-\dfrac{3}{2}$ Giải thích các bước giải: $A=\dfrac{9}{x^2-x+5}\\ =\dfrac{9}{x^2-x+\dfrac{1}{4}+\dfrac{19}{4}}\\ =\dfrac{9}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{19}{4}}\\ A \ \max \Leftrightarrow \left(x-\dfrac{1}{2}\right)^2+\dfrac{19}{4} \ \min\\ \Leftrightarrow x=\dfrac{1}{2}\\ \Rightarrow max_A=\dfrac{9}{\dfrac{19}{4}}=\dfrac{36}{19}\\ B=\dfrac{27-12x}{x^2+9}\\ =\dfrac{(4x^2+36)-(4x^2+12x+9)}{x^2+9}\\ =4-\dfrac{(2x+3)^2}{x^2+9} \le 4$ Dấu “=” xảy ra $\Leftrightarrow x=-\dfrac{3}{2}$ Trả lời
Đáp án:
$max_A= \dfrac{36}{19}\Leftrightarrow x=\dfrac{1}{2}\\ max_B= 4\Leftrightarrow x=-\dfrac{3}{2}$
Giải thích các bước giải:
$A=\dfrac{9}{x^2-x+5}\\ =\dfrac{9}{x^2-x+\dfrac{1}{4}+\dfrac{19}{4}}\\ =\dfrac{9}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{19}{4}}\\ A \ \max \Leftrightarrow \left(x-\dfrac{1}{2}\right)^2+\dfrac{19}{4} \ \min\\ \Leftrightarrow x=\dfrac{1}{2}\\ \Rightarrow max_A=\dfrac{9}{\dfrac{19}{4}}=\dfrac{36}{19}\\ B=\dfrac{27-12x}{x^2+9}\\ =\dfrac{(4x^2+36)-(4x^2+12x+9)}{x^2+9}\\ =4-\dfrac{(2x+3)^2}{x^2+9} \le 4$
Dấu “=” xảy ra $\Leftrightarrow x=-\dfrac{3}{2}$