2003/2004+2004/2005+2005/2003 va 3[ so sanh ] cac ban lam cach ngan gon de hieu nhe
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)
a+b) Xét t/g MAB và t/g MDC có:
MB = MC (gt)
AMB = DMC ( đối đỉnh)
MA = MD (gt)
Do đó, t/g MAB = t/g MDC (c.g.c) (đpcm)
=> AB = CD (2 cạnh tương ứng)
MAB = MDC (2 góc tương ứng)
Mà MAB và MDC là 2 góc ở vị trí so le trong nên AB//CD
t/g ABC = t/g CDA (2 cạnh góc vuông) (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=2\left(x-y\right)-\left(x-y\right)^2\)
\(=\left(x-y\right)\left(2-\left(x-y\right)\right)\)
\(\left(x-y\right)\left(2-x+y\right)\)
\(2x-2y-x^2+2x-y^2\)
\(=2\left(x-y\right)-\left(x-y\right)^2\)
\(=\left(x-y\right)\left(2-\left(x-y\right)\right)\)
\(=\left(x-y\right)\left(2-x+y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Có :\(\left(n+3\right)^3-\left(n-3\right)^3\)
\(=\left[\left(n+3\right)-\left(n-3\right)\right]\left[\left(n+3\right)^2+\left(n+3\right)\left(n-3\right)+\left(n-3\right)^2\right]\)
\(=6\left(3n^2+9\right)==18\left(n^2+3\right)\) chia hết 18 (DPCM)
Ta có:\(\left(n+3\right)^3-\left(n-3\right)^3\)
\(=\left[\left(n+3\right)-\left(n-3\right)\right]\left[\left(n+3\right)^2+\left(n+3\right)\left(n-3\right)+\left(n-3\right)^2\right]\)
\(=\left(n+3-n+3\right)\left(n^2+6n+9+n^2-9+n^2-6n+9\right)\)
\(=6.\left(3n^2+9\right)\)
\(=6.3.\left(n^2+3\right)\)
\(=18.\left(n^2+3\right)⋮18\)
Vậy \(\left(n+3\right)^3-\left(n-3\right)^3⋮18\)
\(\frac{2003}{2004}+\frac{2004}{2005}+\frac{2005}{2003}=1-\frac{1}{2004}+1-\frac{1}{2005}+1+\frac{2}{2003}\)
\(=3+\left(\frac{1}{2003}-\frac{1}{2004}\right)+\left(\frac{1}{2003}-\frac{1}{2005}\right)\)
Do \(\frac{1}{2003}>\frac{1}{2004}>\frac{1}{2005}.\) nên \(\left(\frac{1}{2003}-\frac{1}{2004}\right)+\left(\frac{1}{2003}-\frac{1}{2005}\right)>0\)
Vì vậy \(3+\left(\frac{1}{2003}-\frac{1}{2004}\right)+\left(\frac{1}{2003}-\frac{1}{2005}\right)>3\) (đpcm)
\(A=\frac{2003}{2004}+\frac{2004}{2005}+\frac{2005}{2003}\)
\(=(1-\frac{1}{2004})+(1-\frac{1}{2005})+(1+\frac{2}{2003})\)
\(=3+(\frac{1}{2003}+\frac{1}{2003}-\frac{1}{2004}-\frac{1}{2005})\)
Do\(\frac{1}{2003}\)>\(\frac{1}{2004}\)>\(\frac{1}{2005}\)
\(\Rightarrow\frac{1}{2003}+\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}\)>\(0\)
\(\Rightarrow3+(\frac{1}{2003}-\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2005})\)>\(3\)
\(\Rightarrow A\)>\(3\)