tìm số nguyên x y thỏa mãn (x+3)^2022+(y-2)^2022=0
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\(\left|x-1\right|+\left(y+2\right)^{2022}=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x-1\right|=0\\\left(y+2\right)^{2022}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\\ \Rightarrow B=13.1-5\left(-8\right)+2022=13+40+2022=2075\)
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\(\left(\sqrt{x-1}+\sqrt{3-x}\right)^2\le\left(1^2+1^2\right)\left(x-1+3-x\right)=4\\ \Leftrightarrow\sqrt{x-1}+\sqrt{3-x}\le2\\ y^2+2\sqrt{2020}y+2022=\left(y^2+2y\sqrt{2020}+2020\right)+2\\ =\left(y+\sqrt{2020}\right)^2+2\ge2\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-1=3-x\\y+\sqrt{2020}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-\sqrt{2020}\end{matrix}\right.\)
Vậy ...
ĐKXĐ: \(3\ge x\ge1\)
Áp dụng BĐT Bunhiacopski:
\(1\sqrt{x-1}+1\sqrt{3-x}\le\sqrt{\left(1^2+1^2\right)\left(x-1+3-x\right)}=\sqrt{2.2}=2\)
Mặt khác: \(y^2+2\sqrt{2020}y+2022=\left(y+\sqrt{2020}\right)^2+2\ge2\)
Nên để thõa mãn yêu cầu bài toán thì
\(\left\{{}\begin{matrix}\sqrt{x-1}=\sqrt{3-x}\\y+\sqrt{2020}=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\left(tm\right)\\y=-\sqrt{2020}\end{matrix}\right.\)
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Lời giải:
Gọi $d=ƯCLN(x+2022, x+2015)$
$\Rightarrow (x+2022)-(x+2015)\vdots d$
$\Rightarrow 7\vdots d$
$\Rightarrow d=1$ hoặc $d=7$
Nếu $d=1$ thì $x+2022, x+2015$ nguyên tố cùng nhau
$\Rightarrow (x+2022)^2, (x+2015)^3$ nguyên tố cùng nhau
$\Rightarrow$ để $(x+2022)^2=64(x+2015)^3$ thì:
$x+2015=1, (x+2022)^2=64$
$\Rightarrow x=-2014$ (tm)
Nếu $d=7$ thì đặt $x+2022=7a, x+2015=7b$ với $a,b$ nguyên tố cùng nhau.
Khi đó: $(7a)^2=64(7b)^3$
$\Rightarrow a^2=448b^3$
Vì $(a,b)=1$ nên $b=1; a^2=448$ (vô lý vì 448 không là scp)
Vậy.......
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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 : \(\left|x-2021\right|+\left|y-2022\right|\le0\)
=> \(\left\{{}\begin{matrix}x-2021=0\\y-2022=0\end{matrix}\right.=>\left(x;y\right)=\left(2021;2022\right)\)
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Lời giải:
Ta có:\(y^2+2\sqrt{2020}y+2022=(y^2+2\sqrt{2020}y+2020)+2=(y+\sqrt{2020})^2+2\geq 2(1)\)
Áp dụng BĐT Bunhiacopxky:
$(\sqrt{x-1}+\sqrt{3-x})^2\leq (x-1+3-x)(1+1)=4$
$\Rightarrow \sqrt{x-1}+\sqrt{3-x}\leq 2(2)$
Từ $(1); (2)\Rightarrow \sqrt{x-1}+\sqrt{3-x}\leq 2\leq y^2+2\sqrt{2020}y+2022$
Dấu "=" xảy ra khi mà: \(\left\{\begin{matrix} \frac{x-1}{1}=\frac{3-x}{1}\\ y+\sqrt{2020}=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=2\\ y=-\sqrt{2020}\end{matrix}\right.\)