Tìm GTNN x2 + 2021x 2x^2 – 4x +1 3x^2 – 2x 09/07/2021 Bởi Mackenzie Tìm GTNN x2 + 2021x 2x^2 – 4x +1 3x^2 – 2x
`x^2 + 2021x` `=(4x^2+4.2021x+2021^2-2021^2)/4` `=((2x+2021)^2-2021^2)/4≥(-2021^2)/4` `”=”`xẩy ra khi : `2x+2021=0` `⇔x=(-2021)/2` `2x^2 – 4x +1` `=2(x^2-2x+1)-1` `=2(x-1)^2-1≥-1` `”=”`xẩy ra khi : `x-1=0` `⇔x=1` `3x^2 – 2x` `=(9x^2-6x+1-1)/3` `=((3x-1)^2-1)/3 ≥-1/3` `”=”`xẩy ra khi : `3x-1=0` `⇔x=1/3` Bình luận
Giải thích các bước giải: a) $A=x^2+2021x$ $=x^2+2.x.\dfrac{2021}{2}+\dfrac{4084441}{4}-\dfrac{4084441}{4}$ $=\left(x+\dfrac{2021}{2}\right)^2-\dfrac{4084441}{4}\ge -\dfrac{4084441}{4}$ $A\ge -\dfrac{4084441}{4}⇒A_{min}=-\dfrac{4084441}{4}$ Dấu “=” xảy ra khi: $x+\dfrac{2021}{2}=0$ $⇔x=-\dfrac{2021}{2}$ Vậy $A_{min}=-\dfrac{4084441}{4}$ khi $x=-\dfrac{2021}{2}$ b) $B=2x^2-4x+1$ $=2\left(x^2-2x+\dfrac{1}{2}\right)$ $=2\left(x^2-2x+1-\dfrac{1}{2}\right)$ $=2(x-1)^2-1\ge -1$ $B\ge -1⇒B_{min}=-1$ Dấu “=” xảy ra khi: $x-1=0$ $⇒x=1$ Vậy $B_{min}=-1$ khi $x=1$ c) $C=3x^2-2x$ $=3\left(x^2-\dfrac{2}{3}x\right)$ $=3\left(x^2-2.x.\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{9}\right)$ $=3\left(x-\dfrac{1}{3}\right)^2-\dfrac{1}{3}\ge -\dfrac{1}{3}$ $C\ge -\dfrac{1}{3}⇒C_{min}=-\dfrac{1}{3}$ Dấu “=” xảy ra khi: $x-\dfrac{1}{3}=0$ $⇔x=\dfrac{1}{3}$ Vậy $C_{min}=-\dfrac{1}{3}$ khi $x=\dfrac{1}{3}$. Bình luận
`x^2 + 2021x`
`=(4x^2+4.2021x+2021^2-2021^2)/4`
`=((2x+2021)^2-2021^2)/4≥(-2021^2)/4`
`”=”`xẩy ra khi :
`2x+2021=0`
`⇔x=(-2021)/2`
`2x^2 – 4x +1`
`=2(x^2-2x+1)-1`
`=2(x-1)^2-1≥-1`
`”=”`xẩy ra khi :
`x-1=0`
`⇔x=1`
`3x^2 – 2x`
`=(9x^2-6x+1-1)/3`
`=((3x-1)^2-1)/3 ≥-1/3`
`”=”`xẩy ra khi :
`3x-1=0`
`⇔x=1/3`
Giải thích các bước giải:
a) $A=x^2+2021x$
$=x^2+2.x.\dfrac{2021}{2}+\dfrac{4084441}{4}-\dfrac{4084441}{4}$
$=\left(x+\dfrac{2021}{2}\right)^2-\dfrac{4084441}{4}\ge -\dfrac{4084441}{4}$
$A\ge -\dfrac{4084441}{4}⇒A_{min}=-\dfrac{4084441}{4}$
Dấu “=” xảy ra khi:
$x+\dfrac{2021}{2}=0$
$⇔x=-\dfrac{2021}{2}$
Vậy $A_{min}=-\dfrac{4084441}{4}$ khi $x=-\dfrac{2021}{2}$
b) $B=2x^2-4x+1$
$=2\left(x^2-2x+\dfrac{1}{2}\right)$
$=2\left(x^2-2x+1-\dfrac{1}{2}\right)$
$=2(x-1)^2-1\ge -1$
$B\ge -1⇒B_{min}=-1$
Dấu “=” xảy ra khi:
$x-1=0$
$⇒x=1$
Vậy $B_{min}=-1$ khi $x=1$
c) $C=3x^2-2x$
$=3\left(x^2-\dfrac{2}{3}x\right)$
$=3\left(x^2-2.x.\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{9}\right)$
$=3\left(x-\dfrac{1}{3}\right)^2-\dfrac{1}{3}\ge -\dfrac{1}{3}$
$C\ge -\dfrac{1}{3}⇒C_{min}=-\dfrac{1}{3}$
Dấu “=” xảy ra khi:
$x-\dfrac{1}{3}=0$
$⇔x=\dfrac{1}{3}$
Vậy $C_{min}=-\dfrac{1}{3}$ khi $x=\dfrac{1}{3}$.