Cho x,y ∈ N*, x+y= 2021 Tìm GTNN của P=xy 13/09/2021 Bởi Margaret Cho x,y ∈ N*, x+y= 2021 Tìm GTNN của P=xy
Đáp án: $P\ge -(\dfrac{2019^2}{4}-(\dfrac{2021}2)^2)$ Giải thích các bước giải: Ta có $x, y\in N^*\to 1\le x,y$ Mà $x+y=2021$ $\to x,y\le 2020$ $\to 1\le x, y\le 2021$ Ta có: $x+y=2021$ $\to y=2021-x$ $\to xy=x(2021-x)$ $\to P=2021x-x^2$ $\to -P=x^2-2021x$ $\to -P=(x-\dfrac{2021}{2})^2-(\dfrac{2021}2)^2$ Mà $1\le x\le 2020$ $\to 1-\dfrac{2021}{2}\le x-\dfrac{2021}{2}\le 2020-\dfrac{2021}{2}$ $\to \dfrac{-2019}{2}\le x-\dfrac{2021}{2}\le \dfrac{2019}{2}$ $\to (x-\dfrac{2021}{2})^2\le \dfrac{2019^2}{4}$ $\to (x-\dfrac{2021}{2})^2-(\dfrac{2021}2)^2\le \dfrac{2019^2}{4}-(\dfrac{2021}2)^2$ $\to -P\le \dfrac{2019^2}{4}-(\dfrac{2021}2)^2$ $\to P\ge -(\dfrac{2019^2}{4}-(\dfrac{2021}2)^2)$ Dấu = xảy ra khi $x=2020$ hoặc $x=1$ Bình luận
Đáp án: $P\ge -(\dfrac{2019^2}{4}-(\dfrac{2021}2)^2)$
Giải thích các bước giải:
Ta có $x, y\in N^*\to 1\le x,y$
Mà $x+y=2021$
$\to x,y\le 2020$
$\to 1\le x, y\le 2021$
Ta có:
$x+y=2021$
$\to y=2021-x$
$\to xy=x(2021-x)$
$\to P=2021x-x^2$
$\to -P=x^2-2021x$
$\to -P=(x-\dfrac{2021}{2})^2-(\dfrac{2021}2)^2$
Mà $1\le x\le 2020$
$\to 1-\dfrac{2021}{2}\le x-\dfrac{2021}{2}\le 2020-\dfrac{2021}{2}$
$\to \dfrac{-2019}{2}\le x-\dfrac{2021}{2}\le \dfrac{2019}{2}$
$\to (x-\dfrac{2021}{2})^2\le \dfrac{2019^2}{4}$
$\to (x-\dfrac{2021}{2})^2-(\dfrac{2021}2)^2\le \dfrac{2019^2}{4}-(\dfrac{2021}2)^2$
$\to -P\le \dfrac{2019^2}{4}-(\dfrac{2021}2)^2$
$\to P\ge -(\dfrac{2019^2}{4}-(\dfrac{2021}2)^2)$
Dấu = xảy ra khi $x=2020$ hoặc $x=1$