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![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(M=2022-\left|x-9\right|\le2022\)
\(maxM=2022\Leftrightarrow x=9\)
b) \(N=\left|x-2021\right|+2022\ge2022\)
\(minN=2022\Leftrightarrow x=2021\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\left(3+x\right)^{2022}+\left|2y-1\right|-5\ge-5\\ P_{min}=-5\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=\dfrac{1}{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\left|x-2021\right|+\left|2022-x\right|\ge\left|x-2021+2022-x\right|=1\\ M_{min}=1\Leftrightarrow\left(x-2021\right)\left(2022-x\right)\ge0\Leftrightarrow2021\le x\le2022\)
![](https://rs.olm.vn/images/avt/0.png?1311)
M=|x-2|+|2022-x|>=|x-2+2022-x|=2020
Dấu = xảy ra khi 2<=x<=2022
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a=2022.\left|x^2+1\right|+2023\)
\(\Rightarrow a=2022.\left(x^2+1\right)+2023\left(\left|x^2+1\right|>0,\forall x\right)\)
mà \(\left(x^2+1\right)\ge1,\forall x\)
\(\Rightarrow a=2022.\left(x^2+1\right)+2023\ge2022.1+2023=4045\)
\(\Rightarrow GTNN\left(a\right)=4045\left(x=0\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = (x+5)2022 + | y - 2021| + 2022
vì ( x+5)2022 \(\ge\) 0;
|y-2021| \(\ge\) 0
2022 = 2022
Cộng vế với vế ta được : A = (x+5)2022+|y-2021|+2022\(\ge\) 2022
Vậy A(min) = 2022 dấu bằng xảy ra khi : \(\left\{{}\begin{matrix}x+5=0\\y-2021=0\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x=-5\\y=2021\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\left(2x-y+7\right)^{2022}>=0\forall x,y\)
\(\left|x-1\right|^{2023}>=0\forall x\)
=>\(\left(2x-y+7\right)^{2022}+\left|x-1\right|^{2023}>=0\forall x,y\)
mà \(\left(2x-y+7\right)^{2022}+\left|x-1\right|^{2023}< =0\forall x,y\)
nên \(\left(2x-y+7\right)^{2022}+\left|x-1\right|^{2023}=0\)
=>\(\left\{{}\begin{matrix}2x-y+7=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2x+7=9\end{matrix}\right.\)
\(P=x^{2023}+\left(y-10\right)^{2023}\)
\(=1^{2023}+\left(9-10\right)^{2023}\)
=1-1
=0
c: \(\left|x-3\right|>=0\forall x\)
=>\(\left|x-3\right|+2>=2\forall x\)
=>\(\left(\left|x-3\right|+2\right)^2>=4\forall x\)
mà \(\left|y+3\right|>=0\forall y\)
nên \(\left(\left|x-3\right|+2\right)^2+\left|y+3\right|>=4\forall x,y\)
=>\(P=\left(\left|x-3\right|+2\right)^2+\left|y-3\right|+2019>=4+2019=2023\forall x,y\)
Dấu '=' xảy ra khi x-3=0 và y-3=0
=>x=3 và y=3
Ta có:\(\left|x-1\right|\ge0\)
\(\left|x-2022\right|\ge0\)
=> \(Min_C=0\)
có cách nào dễ hiểu dành cho học sinh lớp 7 ko anh