tìm x,y thỏa mãn:|x-2021|+|y-2022| nhỏ hơn hoặc bằng 0
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a. Nếu $x\geq 3$ thì:
$5=|x-3|-|2-x|=(x-3)-(x-2)=-1$ (vô lý - loại)
Nếu $2\leq x<3$ thì:
$5=|x-3|-|2-x|=(3-x)-(x-2)=5-2x$
$\Rightarrow x=0$ (vô lý - loại do $x\geq 2$)
Nếu $x<2$ thì:
$5=|x-3|-|2-x|=(3-x)-(2-x)=1$ (vô lý - loại)
Vậy không tồn tại $x$ thỏa đề.
b.
Vì $|3x+1|\geq 0; |x-4|\geq 0$ với mọi $x$
Do đó để tổng của chúng bằng $0$ thì:
$|3x+1|=|x-4|=0$
Hay $x=\frac{-1}{3}=4$ (vô lý)
Vậy không tìm được $x$ thỏa mãn.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-1\right)^2+\left(y+3\right)^2=0\left(1\right)\)
Ta thấy \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0,\forall x\\\left(y+3\right)^2\ge0,\forall y\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left(x+1\right)^2=0^2\\\left(y+3\right)^2=0^2\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x+1=0\\y+3=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Áp dụng BĐT $|a|+|b|\geq |a+b|$ ta có:
$|x+1|+|x+5|=|x+1|+|-x-5|\geq |x+1+(-x-5)|=4$
$\Rightarrow |x+1|+|x+5|=3$ là vô lý
Vậy không tìm được $x$ thỏa mãn.
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$N=(a-2)(a+3)-(a-3)(a+2)=(a^2+a-6)-(a^2-a-6)=2a$ không có cơ sở để khẳng định đó là bội của $50$ bạn nhé.
![](https://rs.olm.vn/images/avt/0.png?1311)
= \(\left(\dfrac{2}{3}\right)^3-4.\left(-\dfrac{7}{4}\right)^2+\left(-\dfrac{2}{3}\right)^3\)
= \(\dfrac{8}{27}-4.\dfrac{49}{16}+\left(-\dfrac{8}{27}\right)\)
= \(\left[\dfrac{8}{27}+\left(-\dfrac{8}{27}\right)\right]-4.\dfrac{49}{16}\)
= \(-\dfrac{49}{4}\)
\(\left(\dfrac{2}{3}\right)^3-4.\left(-1\dfrac{3}{4}\right)^2+\left(\dfrac{-2}{3}\right)^3\)
= \(\left[\left(\dfrac{2}{3}\right)^3+\left(\dfrac{-2}{3}\right)^3\right]-4.\left(-1\dfrac{3}{4}\right)^2\)
= \(\left(\dfrac{8}{9}+\dfrac{-8}{9}\right)-4.\left(-1\dfrac{3}{4}\right)^2\)
= \(0-4.\left(\dfrac{-7}{4}\right)^2\)
= \(0-4.\dfrac{49}{16}\)
= \(0-\dfrac{49}{4}\)
= \(\dfrac{-49}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=-\left(2x+1\right)^2-7\left(y-3,5\right)^2+\dfrac{2}{3}\)
vì \(\left\{{}\begin{matrix}-\left(2x+1\right)^2\le0,\forall x\\-7\left(y-3,5\right)^2\le0,\forall y\end{matrix}\right.\)
\(\Rightarrow P=-\left(2x+1\right)^2-7\left(y-3,5\right)^2+\dfrac{2}{3}\le\dfrac{2}{3}\)
Dấu "=" xảy ra khi
\(\left\{{}\begin{matrix}2x+1=0\\y-3,5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=3,5=\dfrac{7}{2}\end{matrix}\right.\)
Vậy \(GTLN\left(P\right)=\dfrac{2}{3}\left(tạix=-\dfrac{1}{2};y=\dfrac{7}{2}\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=\dfrac{5}{3-\left(4x+1\right)^2}\)
Điều kiện xác định khi
\(3-\left(4x+1\right)^2\ne0\Leftrightarrow\left[{}\begin{matrix}4x+1\ne\sqrt[]{3}\\4x+1\ne-\sqrt[]{3}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ne\dfrac{\sqrt[]{3}-1}{4}\\x\ne\dfrac{-\sqrt[]{3}-1}{4}\end{matrix}\right.\)
Ta có :
\(\left(4x+1\right)^2\ge0,\forall x\)
\(\Leftrightarrow3-\left(4x+1\right)^2\le3\)
\(\Leftrightarrow C=\dfrac{5}{3-\left(4x+1\right)^2}\ge\dfrac{5}{3}\)
Vậy \(GTNN\left(C\right)=\dfrac{5}{3}\left(tạix=-\dfrac{1}{4}\right)\)
\(B=\left(2x\right)^2+2\left(y-1\right)^2-5\)
vì \(\left\{{}\begin{matrix}\left(2x\right)^2\ge0,\forall x\\2\left(y-1\right)^2\ge0,\forall y\end{matrix}\right.\)
\(\Rightarrow B=\left(2x\right)^2+2\left(y-1\right)^2-5\ge-5\)
Dấu "=" xảy tại khi
\(\left\{{}\begin{matrix}2x=0\\2\left(y-1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\)
Vậy \(GTNN\left(B\right)=-5\left(tạix=0;y=1\right)\)
Ta thấy : \(\left|x-2021\right|\ge0\forall x,\left|y-2022\right|\ge0\forall y\\ =>\left|x-2021\right|+\left|y-2022\right|\ge0\)
Mà theo đề : \(\left|x-2021\right|+\left|y-2022\right|\le0\)
=> \(\left\{{}\begin{matrix}x-2021=0\\y-2022=0\end{matrix}\right.=>\left(x;y\right)=\left(2021;2022\right)\)