\(\frac{2003}{2004}+\frac{2004}{2005}+\frac{2005}{2003}=1-\frac{1}{2004}+1-\frac{1}{2005}+1+\frac{2}{2003}\)

\(=3+\left(\frac{1}{2003}-\frac{1}{2004}\right)+\left(\frac{1}{2003}-\frac{1}{2005}\right)\)

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\(\frac{1}{2003}\)>\(\frac{1}{2004}\)>\(\frac{1}{2005}\)

\(\Rightarrow\frac{1}{2003}+\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}\)>\(0\)

\(\Rightarrow3+(\frac{1}{2003}-\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2005})\)>\(3\)

\(\Rightarrow A\)>\(3\)

a.Vì OM vuông góc OM; OB vuông góc ON nên \(\widehat{AOM}=\widehat{BON}=90^0\)

b. ta có \(\widehat{NOA}+\widehat{MON}=90^0\left(gt\right);\widehat{MOB}+\widehat{MON}=90^0\left(gt\right)\)

Do vậy \(\widehat{NOA}=\widehat{MOB}\)

\(S=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{2019.2020}\)

\(=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+...\left(\frac{1}{2019}-\frac{1}{2020}\right)\)

\(=1-\frac{1}{2020}=\frac{2019}{2020}\)

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)

15/47<21/65

\(\frac{15}{47}< \frac{21}{65}\)

tt với Ziiz nha

\(=2\left(x-y\right)-\left(x-y\right)^2\)

\(=\left(x-y\right)\left(2-\left(x-y\right)\right)\)

\(\left(x-y\right)\left(2-x+y\right)\)

\(2x-2y-x^2+2x-y^2\)

\(=2\left(x-y\right)-\left(x-y\right)^2\)

\(=\left(x-y\right)\left(2-\left(x-y\right)\right)\)

\(=\left(x-y\right)\left(2-x+y\right)\)

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 \(\left(n+3\right)^3-\left(n-3\right)^3⋮18\)