![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
=> (xy).(yz).(zx) = z. (4x).(9y)
=> (xyz)2 = 36.(xyz)
=> (xyz)2 - 36.(xyz) = 0
=> (xyz).(xyz - 36) = 0
=> xyz = 0 hoặc xyz - 36 = 0
+) xyz = 0 .kết hợp bài cho => x = y = z = 0
+) xyz - 36 = 0 => xyz = 36 mà xy = z nên z.z = 36 => z = 6
Ta có yz = 4x => xyz = x.4x = 36 => x.x = 9 => x = 3
=> y = 36 : xz = 36 : 18 = 2
Vậy....
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{y.\left(y+1\right)}=\frac{2008}{4002}\)
\(=2.\left(\frac{1}{2}-\frac{1}{3}+...+\frac{1}{y}-\frac{1}{y+1}\right)=\frac{1999}{2002}\)
\(=2.\left(\frac{1}{2}-\frac{1}{y+1}\right)=\frac{1999}{2001}\)
\(\frac{1}{2}-\frac{1}{y+1}=\frac{1999}{2001}:2\)
\(\frac{1}{y+1}=\frac{1}{2}-\frac{1999}{4002}\)
\(\frac{1}{y+1}=\frac{1}{2001}\)
\(\Rightarrow y+1=2001\Rightarrow y=2000\)
Ta có: 1/3+1/6+1/10+...+2/y×(y+1)=1999/2001
2/6+2/12+2/20+...+2/y×(y+1)=1999/2001
2×(1/2×3+1/3×4+1/4×5+...+2/y×(y+1)=1999/2001
2×(1/2-1/3+1/3-1/4+1/4-1/5+...+1/y-1/y+1=1999/2001
(1/2-1/y+1)=1999/2001 : 2
1/2-1/y+1=1999/4002
1/y+1=1/2-1999/4002
1/y+1=1/2001
=>y+1=2001
=>y=2001
Có \(1999\)lẻ
\(\Rightarrow x^2+y^2\)lẻ
=> \(x\)hoặc \(y\)lẻ
Giả sử x lẻ , y chẵn
\(\Rightarrow\hept{\begin{cases}x=2m+1\\y=2n\end{cases}}\)
\(\Rightarrow x^2+y^2=\left(2m+1\right)^2+\left(2n\right)^2=4m^2+4m+1+4n^2=1999\)
Do \(4m^2+4m+4n+1\)chia 4 dư 1 mà 1999 chia 4 dư 3 nên ko có x , y tm
Vậy không có x , y tm bài toán