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Đặt \(n^2+2021=k^2\left(k\in N\right)\)
\(\Rightarrow k^2-n^2=2021\\ \Rightarrow\left(k-n\right)\left(k+n\right)=2021\)
Mà \(k,n\in N\)
\(\Rightarrow\left(k-n\right)\left(k+n\right)=2021\cdot1=43\cdot47\)
\(\left\{{}\begin{matrix}k-n=2021\\k+n=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}k=1011\\n=-1010\left(loại\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}k-n=1\\k+n=2021\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}k=1011\\n=1010\end{matrix}\right.\left(nhận\right)\)
\(\left\{{}\begin{matrix}k-n=43\\k+n=47\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}k=45\\n=2\end{matrix}\right.\left(nhận\right)\)
\(\left\{{}\begin{matrix}k-n=47\\k+n=43\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}k=45\\n=-2\left(loại\right)\end{matrix}\right.\)
Vậy \(n\in\left\{2;1010\right\}\)
Giả sử n2+2021 là SCP
\(Đặtn^2+2021=k^2\left(k\in N\right)\\ \Rightarrow n^2-k^2=-2021\\ \Rightarrow\left(n-k\right)\left(n+k\right)=-2021\)
Vì \(n,k\in N\Rightarrow\left\{{}\begin{matrix}n-k< n+k\\n-k,n+k\in Z\\n-k,n+k\inƯ\left(-2021\right)\end{matrix}\right.\)
Ta có bảng:
n-k | -43 | -47 |
n+k | 47 | 43 |
n | 2 | -2 |
Mà n∈N⇒n=2
Vậy n=2
![](https://rs.olm.vn/images/avt/0.png?1311)
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Đk:\(x\ge3;y\ge2021\)
\(A=x+y-\sqrt{x-3}.\sqrt{y-2021}\)
\(\Leftrightarrow A=\left(x-3\right)-\sqrt{x-3}.\sqrt{y-2021}+\dfrac{1}{4}\left(y-2021\right)+\dfrac{3}{4}\left(y-2021\right)+2024\)
\(\Leftrightarrow A=\left(\sqrt{x-3}-\dfrac{1}{2}\sqrt{y-2021}\right)^2+\dfrac{3}{4}\left(y-2021\right)+2024\ge2024\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}y-2021=0\\\sqrt{x-3}-\dfrac{1}{2}\sqrt{y-2021}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}y=2021\\x=3\end{matrix}\right.\) (tm)
Vậy...
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+) \(\left(x^2+1\right)\left(x+1\right)=4^y\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=2^{2y}\)
+) Do \(x,y\inℕ\)nên ta có \(x^2+1=2^m\)và \(x+1=2^n\)với \(m+n=2y;m,n\inℕ\)
+) Lúc đó ta có: \(\orbr{\begin{cases}x^2+1⋮x+1\\x+1⋮x^2+1\end{cases}}\)
TH1: \(x^2+1⋮x+1\Leftrightarrow\left(x+1\right)^2-2\left(x+1\right)+2⋮x+1\)
\(\Leftrightarrow2⋮x+1\Leftrightarrow x\in\left\{0;1\right\}\)
TH2: \(x+1⋮x^2+1\Leftrightarrow x^2-1⋮x^2+1\Leftrightarrow2⋮x+1\)
\(\Leftrightarrow x\in\left\{0;1\right\}\)
* Nếu x = 0 thì \(4^y=1\Leftrightarrow y=0\)
* Nếu y = 0 thì \(4^y=4\Leftrightarrow y=1\)
Vậy \(\left(x;y\right)\in\left\{\left(0;0\right);\left(1;1\right)\right\}\)
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(2n + 1)(y - 3) = 10 = 1.10 = 10.1 = 2.5 = 5.2
2n + 1 = 1 => n = 0 ; y - 3 = 10 => y = 13
2n + 1 = 10 => n = 4,5 (loại)
2n + 1 = 2 => n = 0,5 (Loại)
2n + 1 = 5 => n =2 ; y - 3 = 2 => y = 5
Vậy các cặp (x;y) là (0;13) ; (2;5)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow x+2\sqrt{3}=y+z+2\sqrt{yz}\)
\(\Leftrightarrow x-y-z=2\left(\sqrt{yz}-\sqrt{3}\right)\)
Do x;y;z;2 đều là các số hữu tỉ mà \(\sqrt{yz}-\sqrt{3}\) vô tỉ
Nên đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}x-y-z=0\\yz=3\end{matrix}\right.\)
\(\Leftrightarrow\left(x;y;z\right)=\left(4;3;1\right);\left(4;1;3\right)\)
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+) x = 0
=> 1 + 2123 = 708y
<=> y = 3
+) x > 0
=> \(2123=708y-15^x⋮3\) vô lí vì 2123 không chia hết cho 3
Vậy x = 0 và y = 3
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\(x^3+y^3+3xy\left(x+y\right)+\dfrac{1}{27}-3xy\left(x+y\right)-xy=0\)
\(\Leftrightarrow\left(x+y\right)^3+\dfrac{1}{27}-3xy\left(x+y+\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow\left(x+y+\dfrac{1}{3}\right)\left[\left(x+y\right)^2-\dfrac{1}{3}\left(x+y\right)+\dfrac{1}{9}\right]-3xy\left(x+y+\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow x^2+y^2-xy-\dfrac{1}{3}\left(x+y\right)+\dfrac{1}{9}=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(x-\dfrac{1}{3}\right)^2+\left(y-\dfrac{1}{3}\right)^2=0\)
\(\Leftrightarrow x=y=\dfrac{1}{3}\Rightarrow P=...\)