Toán Timg giá trị lớn nhất của biểu thức: A= 4a – 2a^2 B= -2y^2 – 10y + 15 06/08/2021 By Ivy Timg giá trị lớn nhất của biểu thức: A= 4a – 2a^2 B= -2y^2 – 10y + 15
Đáp án: $A = 4a-2a^2$ $ = -(2a^2-4a)$ $ = -[(√2a)^2 – 2 .√2a .√2 +2-2]$ $ = -(√2a-√2)^2 +2$ Vì $-(√2a-√2)^2 ≤ 0$ Nên $-(√2a-√2)^2 +2 ≤ 2$ Dấu ”=” xảy ra khi $√2a-√2 =0⇔a=1$ Vậy Max A $=2$ , tại $a=1$ $B = -2y^2-10y+15$ $ = -(2y² +10y-15)$ $ = -[(√2y)^2 + 2 .√2y. \dfrac{5\sqrt[]{2}}{2} + \dfrac{25}{2} -\dfrac{55}{2}]$ $ = -(√2y + \dfrac{5\sqrt[]{2}}{2})^2 + \dfrac{55}{2}$ Vì $-(√2y +\dfrac{5\sqrt[]{2}}{2})^2 ≤ 0$ Nên $-(√2y +\dfrac{5\sqrt[]{2}}{2})^2 +\dfrac{55}{2} ≤ \dfrac{55}{2}$ Dấu ”=” xảy ra khi $√2y +\dfrac{5\sqrt[]{2}}{2} = 0⇔y =-\dfrac{5}{2}$ Vậy Max B $=\dfrac{55}{2}$ , tại $y=-\dfrac{5}{2}$ Trả lời
$Ta có:A=4a-2a^{2}$ $=4a-2a^{2} -2+2$ $=2-(2a^{2}-4a+2)$ $=2-2(a^{2} -2a+1)$ $=2-2(a-1)^{2} .$ $Vì 2(a-1)^{2} ≥0 ∀x$ $⇔2-2(a-1)^{2} ≤2 ∀x$ $⇔A≤2 ∀x.$ $Dấu\text{“=”}\text {xảy}\text {ra}⇔2(a-1)^{2} =0$ $⇔(a-1)^{2}=0$ $⇔a-1=0$ $⇔a=1.$ $Vậy\text{a}=1\text {thì}\text {Max} A=2.$ $Ta\text {có}:B=-2y^{2} -10y+15$ $=-2y^{2}-10y-\frac{25}{2} +27\frac{1}{2}$ $=27\frac{1}{2} -(2y^{2}+10y+\frac{25}{2}$) $=27\frac{1}{2}-2(y^{2}+5y+\frac{25}{4}$) $=27\frac{1}{2}-2(y+\frac{5}{2} )^{2}$ . $Ta\text{ có}:2(y+\frac{5}{2} )^{2} ≥0∀y$ $⇒27\frac{1}{2}-2(y+\frac{5}{2} )^{2}≤27\frac{1}{2}∀y$ $⇒B≤27\frac{1}{2} ∀y.$ $Dấu\text{“=”}\text{xảy}\text{ra}⇔2(y+\frac{5}{2} )^{2}=0$ $⇔(y+\frac{5}{2} )^{2}=0$ $⇔y+\frac{5}{2}=0$ $⇔y=-\frac{5}{2} .$ $Vậy\text{ y}=-\frac{5}{2}\text{thì}\text {Max}\text{B}=27\frac{1}{2}.$ Trả lời
Đáp án:
$A = 4a-2a^2$
$ = -(2a^2-4a)$
$ = -[(√2a)^2 – 2 .√2a .√2 +2-2]$
$ = -(√2a-√2)^2 +2$
Vì $-(√2a-√2)^2 ≤ 0$
Nên $-(√2a-√2)^2 +2 ≤ 2$
Dấu ”=” xảy ra khi $√2a-√2 =0⇔a=1$
Vậy Max A $=2$ , tại $a=1$
$B = -2y^2-10y+15$
$ = -(2y² +10y-15)$
$ = -[(√2y)^2 + 2 .√2y. \dfrac{5\sqrt[]{2}}{2} + \dfrac{25}{2} -\dfrac{55}{2}]$
$ = -(√2y + \dfrac{5\sqrt[]{2}}{2})^2 + \dfrac{55}{2}$
Vì $-(√2y +\dfrac{5\sqrt[]{2}}{2})^2 ≤ 0$
Nên $-(√2y +\dfrac{5\sqrt[]{2}}{2})^2 +\dfrac{55}{2} ≤ \dfrac{55}{2}$
Dấu ”=” xảy ra khi $√2y +\dfrac{5\sqrt[]{2}}{2} = 0⇔y =-\dfrac{5}{2}$
Vậy Max B $=\dfrac{55}{2}$ , tại $y=-\dfrac{5}{2}$
$Ta có:A=4a-2a^{2}$
$=4a-2a^{2} -2+2$
$=2-(2a^{2}-4a+2)$
$=2-2(a^{2} -2a+1)$
$=2-2(a-1)^{2} .$
$Vì 2(a-1)^{2} ≥0 ∀x$
$⇔2-2(a-1)^{2} ≤2 ∀x$
$⇔A≤2 ∀x.$
$Dấu\text{“=”}\text {xảy}\text {ra}⇔2(a-1)^{2} =0$
$⇔(a-1)^{2}=0$
$⇔a-1=0$
$⇔a=1.$
$Vậy\text{a}=1\text {thì}\text {Max} A=2.$
$Ta\text {có}:B=-2y^{2} -10y+15$
$=-2y^{2}-10y-\frac{25}{2} +27\frac{1}{2}$
$=27\frac{1}{2} -(2y^{2}+10y+\frac{25}{2}$)
$=27\frac{1}{2}-2(y^{2}+5y+\frac{25}{4}$)
$=27\frac{1}{2}-2(y+\frac{5}{2} )^{2}$ .
$Ta\text{ có}:2(y+\frac{5}{2} )^{2} ≥0∀y$
$⇒27\frac{1}{2}-2(y+\frac{5}{2} )^{2}≤27\frac{1}{2}∀y$
$⇒B≤27\frac{1}{2} ∀y.$
$Dấu\text{“=”}\text{xảy}\text{ra}⇔2(y+\frac{5}{2} )^{2}=0$
$⇔(y+\frac{5}{2} )^{2}=0$
$⇔y+\frac{5}{2}=0$
$⇔y=-\frac{5}{2} .$
$Vậy\text{ y}=-\frac{5}{2}\text{thì}\text {Max}\text{B}=27\frac{1}{2}.$