Toán `1) 1/{x^2+x+1}` `2) 3/{4x^2-4x+5}` Đề bài: Tìm Min hoặc Max. 14/11/2021 By Josie `1) 1/{x^2+x+1}` `2) 3/{4x^2-4x+5}` Đề bài: Tìm Min hoặc Max.
Đáp án : `1)Amax=4/3` khi `x=-1/2` `2)Cmax=3/4` khi `x=1/2` Giải thích các bước giải : `1)A=1/(x^2+x+1)` Để `Amax => 1/(x^2+x+1) min => x^2+x+1 min` Đặt `B=x^2+x+1` `<=>B=x^2+2×x×1/2+(1/2)^2-1/4+1` `<=>B=(x+1/2)^2+3/4` Vì `(x+1/2)^2 ≥ 0` `=>Bmin=3/4` `<=>(x+1/2)^2=0` `<=>x+1/2=0` `<=>x=-1/2` `<=>Amax=1/(3/4)=1:3/4=1×4/3=4/3` Vậy `Amax=4/3` khi `x=-1/2` `2)C=3/(4x^2-4x+5)` Để `Cmax => 3/(4x^2-4x+5) max => 4x^2-4x+5 min` Đặt `D=4x^2-4x+5` `<=>D=(2x)^2-2×2x×1+1+4` `<=>D=(2x-1)^2+4` Vì `(2x-1)^2 ≥ 0` `=>Dmin=4` `<=>(2x-1)^2=0` `<=>2x-1=0` `<=>2x=1` `<=>x=1/2` `<=>Cmax=3/4` Vậy `Cmax=3/4` khi `x=1/2` ~Chúc bạn học tốt !!!~ Trả lời
Đáp án: 1) $\max\left(\dfrac{1}{x^2 + x + 1}\right) = \dfrac43 \Leftrightarrow x = -\dfrac12$ 2) $\max \left(\dfrac{3}{4x^2 -4x + 5}\right) = \dfrac34 \Leftrightarrow x = \dfrac12$ Giải thích các bước giải: $\begin{array}{l}1)\quad A = \dfrac{1}{x^2 + x + 1}\\ \to A = \dfrac{1}{x^2 + 2\cdot\dfrac12x + \dfrac14 + \dfrac34}\\ \to A = \dfrac{1}{\left(x + \dfrac12\right)^2 + \dfrac34}\\ \text{Ta có:}\\ \quad \left(x + \dfrac12\right)^2\geq 0\quad \forall x\\ \to \left(x + \dfrac12\right)^2 + \dfrac34 \geq \dfrac34\\ \to \dfrac{1}{\left(x + \dfrac12\right)^2+\dfrac34} \leq \dfrac43\\ \to A \leq \dfrac43\\ \text{Dấu = xảy ra}\,\,\Leftrightarrow \left(x + \dfrac12\right)^2 = 0\Leftrightarrow x = -\dfrac12\\ Vậy \,\,\max\left(\dfrac{1}{x^2 + x + 1}\right) = \dfrac43 \Leftrightarrow x = -\dfrac12\\ 2)\quad B = \dfrac{3}{4x^2 -4x + 5}\\ \to B = \dfrac{3}{4x^2 – 4x + 1 + 4}\\ \to B = \dfrac{3}{(2x – 1)^2 +4}\\ \text{Ta có:}\\ \quad (2x -1)^2 \geq 0\quad \forall x\\ \to (2x-1)^2 + 4 \geq 4\\ \to \dfrac{3}{(2x-1)^2 + 4} \leq \dfrac{3}{4}\\ \to B \leq \dfrac34\\ \text{Dấu = xảy ra}\,\,\Leftrightarrow (2x – 1)^2 = 0 \Leftrightarrow x = \dfrac12\\ Vậy\,\,\max \left(\dfrac{3}{4x^2 -4x + 5}\right) = \dfrac34 \Leftrightarrow x = \dfrac12 \end{array}$ Trả lời
Đáp án :
`1)Amax=4/3` khi `x=-1/2`
`2)Cmax=3/4` khi `x=1/2`
Giải thích các bước giải :
`1)A=1/(x^2+x+1)`
Để `Amax => 1/(x^2+x+1) min => x^2+x+1 min`
Đặt `B=x^2+x+1`
`<=>B=x^2+2×x×1/2+(1/2)^2-1/4+1`
`<=>B=(x+1/2)^2+3/4`
Vì `(x+1/2)^2 ≥ 0`
`=>Bmin=3/4`
`<=>(x+1/2)^2=0`
`<=>x+1/2=0`
`<=>x=-1/2`
`<=>Amax=1/(3/4)=1:3/4=1×4/3=4/3`
Vậy `Amax=4/3` khi `x=-1/2`
`2)C=3/(4x^2-4x+5)`
Để `Cmax => 3/(4x^2-4x+5) max => 4x^2-4x+5 min`
Đặt `D=4x^2-4x+5`
`<=>D=(2x)^2-2×2x×1+1+4`
`<=>D=(2x-1)^2+4`
Vì `(2x-1)^2 ≥ 0`
`=>Dmin=4`
`<=>(2x-1)^2=0`
`<=>2x-1=0`
`<=>2x=1`
`<=>x=1/2`
`<=>Cmax=3/4`
Vậy `Cmax=3/4` khi `x=1/2`
~Chúc bạn học tốt !!!~
Đáp án:
1) $\max\left(\dfrac{1}{x^2 + x + 1}\right) = \dfrac43 \Leftrightarrow x = -\dfrac12$
2) $\max \left(\dfrac{3}{4x^2 -4x + 5}\right) = \dfrac34 \Leftrightarrow x = \dfrac12$
Giải thích các bước giải:
$\begin{array}{l}1)\quad A = \dfrac{1}{x^2 + x + 1}\\ \to A = \dfrac{1}{x^2 + 2\cdot\dfrac12x + \dfrac14 + \dfrac34}\\ \to A = \dfrac{1}{\left(x + \dfrac12\right)^2 + \dfrac34}\\ \text{Ta có:}\\ \quad \left(x + \dfrac12\right)^2\geq 0\quad \forall x\\ \to \left(x + \dfrac12\right)^2 + \dfrac34 \geq \dfrac34\\ \to \dfrac{1}{\left(x + \dfrac12\right)^2+\dfrac34} \leq \dfrac43\\ \to A \leq \dfrac43\\ \text{Dấu = xảy ra}\,\,\Leftrightarrow \left(x + \dfrac12\right)^2 = 0\Leftrightarrow x = -\dfrac12\\ Vậy \,\,\max\left(\dfrac{1}{x^2 + x + 1}\right) = \dfrac43 \Leftrightarrow x = -\dfrac12\\ 2)\quad B = \dfrac{3}{4x^2 -4x + 5}\\ \to B = \dfrac{3}{4x^2 – 4x + 1 + 4}\\ \to B = \dfrac{3}{(2x – 1)^2 +4}\\ \text{Ta có:}\\ \quad (2x -1)^2 \geq 0\quad \forall x\\ \to (2x-1)^2 + 4 \geq 4\\ \to \dfrac{3}{(2x-1)^2 + 4} \leq \dfrac{3}{4}\\ \to B \leq \dfrac34\\ \text{Dấu = xảy ra}\,\,\Leftrightarrow (2x – 1)^2 = 0 \Leftrightarrow x = \dfrac12\\ Vậy\,\,\max \left(\dfrac{3}{4x^2 -4x + 5}\right) = \dfrac34 \Leftrightarrow x = \dfrac12 \end{array}$