Giúp mình với Đề : tìm min 2x^2 – 3x + 1 01/07/2021 Bởi Elliana Giúp mình với Đề : tìm min 2x^2 – 3x + 1
Đáp án+Giải thích các bước giải: `2x^2-3x+1` `=2.(x^2-(3)/(2)x+(1)/(2))` `=2.[x^2-2.(3)/(4).x+(3/4)^2-(3/4)^2+1/2]` `=2.[(x-3/4)^2-(1)/(16)]` `=2.(x-3/4)^2-1/8` Vì `(x-3/4)^2≥0∀x∈RR=>2.(x-3/4)^2≥0∀x∈RR` `=>2(x-3/4)^2-1/8≥-(1)/(8)∀x∈RR` `=>2x^2-3x+1≥-(1)/(8)∀x∈RR` Dấu `=` xảy ra: `⇔x-3/4=0<=>x=3/4` Vậy `2x^2-3x+1` đạt GTNN là `-1/8` khi `x=3/4` Bình luận
`~rai~` \(A=2x^2-3x+1\\\quad=\left(2x^2-3x+\dfrac{9}{8}\right)-\dfrac{1}{8}\\\quad=2\left(x^2-\dfrac{3}{4}+\dfrac{9}{16}\right)-\dfrac{1}{8}\\\quad=2\left(x-\dfrac{3}{4}\right)^2-\dfrac{1}{8}\\\text{Ta có:}2\left(x-\dfrac{3}{4}\right)^2\ge 0\quad\forall x\in\mathbb{R}\\\Leftrightarrow 2\left(x-\dfrac{3}{4}\right)^2-\dfrac{1}{8}\ge -\dfrac{1}{8}\quad\forall x\in\mathbb{R}\\\Leftrightarrow A\ge -\dfrac{1}{8}\quad\forall x\in\mathbb{R}.\\\text{Dấu “=” xảy ra}\Leftrightarrow \left(x-\dfrac{3}{4}\right)^2=0\Leftrightarrow x=\dfrac{3}{4}.\\\text{Vậy}\quad Min_{A}=-\dfrac{1}{8}\quad khi\quad x=\dfrac{3}{4}.\) Bình luận
Đáp án+Giải thích các bước giải:
`2x^2-3x+1`
`=2.(x^2-(3)/(2)x+(1)/(2))`
`=2.[x^2-2.(3)/(4).x+(3/4)^2-(3/4)^2+1/2]`
`=2.[(x-3/4)^2-(1)/(16)]`
`=2.(x-3/4)^2-1/8`
Vì `(x-3/4)^2≥0∀x∈RR=>2.(x-3/4)^2≥0∀x∈RR`
`=>2(x-3/4)^2-1/8≥-(1)/(8)∀x∈RR`
`=>2x^2-3x+1≥-(1)/(8)∀x∈RR`
Dấu `=` xảy ra:
`⇔x-3/4=0<=>x=3/4`
Vậy `2x^2-3x+1` đạt GTNN là `-1/8` khi `x=3/4`
`~rai~`
\(A=2x^2-3x+1\\\quad=\left(2x^2-3x+\dfrac{9}{8}\right)-\dfrac{1}{8}\\\quad=2\left(x^2-\dfrac{3}{4}+\dfrac{9}{16}\right)-\dfrac{1}{8}\\\quad=2\left(x-\dfrac{3}{4}\right)^2-\dfrac{1}{8}\\\text{Ta có:}2\left(x-\dfrac{3}{4}\right)^2\ge 0\quad\forall x\in\mathbb{R}\\\Leftrightarrow 2\left(x-\dfrac{3}{4}\right)^2-\dfrac{1}{8}\ge -\dfrac{1}{8}\quad\forall x\in\mathbb{R}\\\Leftrightarrow A\ge -\dfrac{1}{8}\quad\forall x\in\mathbb{R}.\\\text{Dấu “=” xảy ra}\Leftrightarrow \left(x-\dfrac{3}{4}\right)^2=0\Leftrightarrow x=\dfrac{3}{4}.\\\text{Vậy}\quad Min_{A}=-\dfrac{1}{8}\quad khi\quad x=\dfrac{3}{4}.\)