Câu hỏi của Phạm Bảo Ngọc

Question

Tìm GTLN và GTNN của

N=$\frac{4x-8}{x^{2\:}-4x+8}$

Nguyễn Việt Lâm · Accepted Answer

$N=\frac{x^2-4x+8-x^2-8x-16}{x^2-4x+8}=1-\frac{\left(x+4\right)^2}{\left(x-2\right)^2+4}\le1$

$N_{max}=1$ khi $x=-4$

$N=\frac{-x^2+4x-8+x^2}{x^2-4x+8}=-1+\frac{x^2}{\left(x-2\right)^2+4}\ge-1$

$N_{min}=-1$ khi $x=0$Câu trả lời: $N=\frac{x^2-4x+8-x^2-8x-16}{x^2-4x+8}=1-\frac{\left(x+4\right)^2}{\left(x-2\right)^2+4}\le1$

$N_{max}=1$ khi $x=-4$

$N=\frac{-x^2+4x-8+x^2}{x^2-4x+8}=-1+\frac{x^2}{\left(x-2\right)^2+4}\ge-1$

$N_{min}=-1$ khi $x=0$

Akai Haruma · Answer

Lời giải:
$N=\frac{4(x-2)}{(x^2-4x+4)+4}=\frac{4(x-2)}{(x-2)^2+4}=\frac{4t}{t^2+4}$

Có:

$N+2=\frac{t^2+4t+4}{t^2+4}=\frac{(t+2)^2}{t^2+4}\geq 0, \forall t\in\mathbb{R}$

$\Rightarrow N\geq -2$ hay $N_{\min}=-2$ khi $t=-2\Leftrightarrow x=0$

$N-2=-\frac{t^2-4t+4}{t^2+4}=\frac{-(t-2)^2}{t^2+4}\leq 0, \forall t\in\mathbb{R}$

$\Rightarrow N\leq 2$ hay $N_{\max}=2$ khi $t=2\Leftrightarrow x=4$

Vậy......Câu trả lời: Lời giải:
$N=\frac{4(x-2)}{(x^2-4x+4)+4}=\frac{4(x-2)}{(x-2)^2+4}=\frac{4t}{t^2+4}$

Có:

$N+2=\frac{t^2+4t+4}{t^2+4}=\frac{(t+2)^2}{t^2+4}\geq 0, \forall t\in\mathbb{R}$

$\Rightarrow N\geq -2$ hay $N_{\min}=-2$ khi $t=-2\Leftrightarrow x=0$

$N-2=-\frac{t^2-4t+4}{t^2+4}=\frac{-(t-2)^2}{t^2+4}\leq 0, \forall t\in\mathbb{R}$

$\Rightarrow N\leq 2$ hay $N_{\max}=2$ khi $t=2\Leftrightarrow x=4$

Vậy......

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