tìm gtnn a=40^2-12a+1 mk cần gấp mong giúp ạ 19/08/2021 Bởi Alexandra tìm gtnn a=40^2-12a+1 mk cần gấp mong giúp ạ
`a=40a^2-12a+1` `⇒a/40 = a^2 – 3/10a + 1/40` `⇔a/40 = (a^2- 2. a . 3/20 + 9 / 400 ) + 1/400` `⇔a/40= (a- 3/20)^2 + 1/400` `⇔a = 40(a-3/20)^2 + 40/400` `⇔a = 40(a-3/20)^2 + 1/10.` Có: `40(a-3/20)^2\ge0⇒40(a-3/20)^2 + 1/10\ge1/10.` Dấu ”=” xảy ra khi `a-3/20=0⇔a=3/20.` Vậy $Min_a$`=1/10⇔a=3/20.` Bình luận
$A=40a^2-12a+1$ $=(2\sqrt{10} a)^2-2.2\sqrt{10}a.\dfrac{3}{\sqrt{10}}+\dfrac{9}{10}+\dfrac{1}{10}$ $=(2\sqrt{10}a-\dfrac{3}{\sqrt{10}})^2+\dfrac{1}{10}\ge \dfrac{1}{10}$ $\min A=\dfrac{1}{10}\Leftrightarrow a=\dfrac{3}{20}$ Bình luận
`a=40a^2-12a+1`
`⇒a/40 = a^2 – 3/10a + 1/40`
`⇔a/40 = (a^2- 2. a . 3/20 + 9 / 400 ) + 1/400`
`⇔a/40= (a- 3/20)^2 + 1/400`
`⇔a = 40(a-3/20)^2 + 40/400`
`⇔a = 40(a-3/20)^2 + 1/10.`
Có: `40(a-3/20)^2\ge0⇒40(a-3/20)^2 + 1/10\ge1/10.`
Dấu ”=” xảy ra khi `a-3/20=0⇔a=3/20.`
Vậy $Min_a$`=1/10⇔a=3/20.`
$A=40a^2-12a+1$
$=(2\sqrt{10} a)^2-2.2\sqrt{10}a.\dfrac{3}{\sqrt{10}}+\dfrac{9}{10}+\dfrac{1}{10}$
$=(2\sqrt{10}a-\dfrac{3}{\sqrt{10}})^2+\dfrac{1}{10}\ge \dfrac{1}{10}$
$\min A=\dfrac{1}{10}\Leftrightarrow a=\dfrac{3}{20}$