Tìm giá trị lớn nhất: D=-5-{x-1}{x+2} E =11-10x- x mũ 2 13/07/2021 Bởi Everleigh Tìm giá trị lớn nhất: D=-5-{x-1}{x+2} E =11-10x- x mũ 2
` D = -5 – (x-1)(x+2)` ` =-5 – (x^2 +2x – x -2)` ` = -5 – ( x^2 + x -2)` ` = -5 – x^2 – x +2` ` = -x^2 – x -3` ` = -(x^2 +x) -3 = – (x^2 + 2 . x . 1/2 + 1/4) + 1/4 – 3` ` = -(x +1/2)^2 -11/4` Ta có ` (x +1/2)^2 \ge 0 \to – (x+1/2)^2 \le 0 \to -(x +1/2)^2 -11/4 \le -11/4` Vậy GTLN của ` D =-11/4` khi ` x +1/2 = 0 \to x= -1/2` —- ` E = 11 – 10x – x^2` ` = – (x^2 +10x ) +11` ` = – (x^2 + 2. 5 . x + 25) + 25 +11` ` = – (x+5)^2 + 36` Ta có ` (x+5)^2 \ge 0 \to – (x+5)^2 \le 0 \to – (x+5)^2 + 36 \le 36` Vậy GTLN của ` E= 36` khi ` x+5 = 0 \to x = -5` Bình luận
$D=-5-(x-1)(x+2)\\=-5-(x^2+x-2)\\=-5-(x^2+x+\dfrac{1}{4})+\dfrac{9}{4}\\=\dfrac{-11}{4}-(x+\dfrac{1}{2})^2$ Vì $(x+\dfrac{1}{2})^2\ge0\forall x\in R$ nên $D\le\dfrac{-11}{4}$ Vậy $D_{max}=\dfrac{-11}{4}$ khi $(x+\dfrac{1}{2})=0$ hay $x=\dfrac{-1}{2}$ $E=11-10x-x^2\\=36-(25+10x+x^2)\\=36-(5+x)^2$ Vì $(5+x)^2\ge0\forall x\in R$ nên $E\le36$ Vậy $E_{max}=36$ khi $5+x=0$ hay $x=-5$ Bình luận
` D = -5 – (x-1)(x+2)`
` =-5 – (x^2 +2x – x -2)`
` = -5 – ( x^2 + x -2)`
` = -5 – x^2 – x +2`
` = -x^2 – x -3`
` = -(x^2 +x) -3 = – (x^2 + 2 . x . 1/2 + 1/4) + 1/4 – 3`
` = -(x +1/2)^2 -11/4`
Ta có
` (x +1/2)^2 \ge 0 \to – (x+1/2)^2 \le 0 \to -(x +1/2)^2 -11/4 \le -11/4`
Vậy GTLN của ` D =-11/4` khi ` x +1/2 = 0 \to x= -1/2`
—-
` E = 11 – 10x – x^2`
` = – (x^2 +10x ) +11`
` = – (x^2 + 2. 5 . x + 25) + 25 +11`
` = – (x+5)^2 + 36`
Ta có
` (x+5)^2 \ge 0 \to – (x+5)^2 \le 0 \to – (x+5)^2 + 36 \le 36`
Vậy GTLN của ` E= 36` khi ` x+5 = 0 \to x = -5`
$D=-5-(x-1)(x+2)\\=-5-(x^2+x-2)\\=-5-(x^2+x+\dfrac{1}{4})+\dfrac{9}{4}\\=\dfrac{-11}{4}-(x+\dfrac{1}{2})^2$
Vì $(x+\dfrac{1}{2})^2\ge0\forall x\in R$ nên $D\le\dfrac{-11}{4}$
Vậy $D_{max}=\dfrac{-11}{4}$ khi $(x+\dfrac{1}{2})=0$ hay $x=\dfrac{-1}{2}$
$E=11-10x-x^2\\=36-(25+10x+x^2)\\=36-(5+x)^2$
Vì $(5+x)^2\ge0\forall x\in R$ nên $E\le36$
Vậy $E_{max}=36$ khi $5+x=0$ hay $x=-5$