a) \(A=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{98\cdot99\cdot100}\)

\(A=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\)

\(A=\frac{1}{2}-\frac{1}{99\cdot100}=\frac{1}{2}-\frac{1}{9900}=\frac{4949}{9900}\)

b) \(B=\frac{17}{1\cdot3\cdot5}+\frac{17}{3\cdot5\cdot7}+\frac{17}{5\cdot7\cdot9}+...+\frac{17}{47\cdot49\cdot51}\)

\(B=\frac{17}{4}\left(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+...+\frac{4}{47\cdot49\cdot51}\right)\)

\(B=\frac{17}{4}\left(\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{47\cdot49}-\frac{1}{49\cdot51}\right)\)

\(B=\frac{17}{4}\left(\frac{1}{3}-\frac{1}{2499}\right)=\frac{17}{4}\cdot\frac{832}{2499}=\frac{208}{147}\)

a/ Ta có

^ADB = ^BEC (Đề bài) (1)

Xét tg AEC có

^BEC = ^A + ^ACE (trong 1 tg góc ngoài bằng tổng 2 góc trong ko kề với nó) (2)

Xét tg BCD có

^ADB = ^ACB + ^DBC (lý do như trên) (3)

Từ (1) (2) và (3) => ^A + ^ACE = ^ACB + ^DBC (dpcm)

b/

Từ KQ của câu A ta có

\(\widehat{A}+\frac{\widehat{C}}{2}=\widehat{C}+\frac{\widehat{B}}{2}\Rightarrow\widehat{A}=\frac{\widehat{B}+\widehat{C}}{2}\Rightarrow2.\widehat{A}=\widehat{B}+\widehat{C}\)

Mà \(\widehat{A}=180^o-\left(\widehat{B}+\widehat{C}\right)\Rightarrow\widehat{A}=180^o-2.\widehat{A}\Rightarrow3.\widehat{A}=180^o\Rightarrow\widehat{A}=60^o\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

Do đó \(\frac{a+b}{a-b}=\frac{bk+b}{bk-b}=\frac{b\left(k+1\right)}{b\left(k-1\right)}=\frac{k+1}{k-1}\)(1)

\(\frac{c+d}{c-d}=\frac{dk+d}{dk-d}=\frac{d\left(k+1\right)}{d\left(k-1\right)}=\frac{k+1}{k-1}\)(2)

Từ (1) và (2) => \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)

b) \(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\right)\frac{2-\left(1+3+5+7+..+49\right)}{12}\)

\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\frac{2-\left(12.50+25\right)}{89}=-\frac{5.9.7.89}{5.4.7.7.89}=\frac{-9}{28}\)

\(\frac{a^2+b^2}{b^2+c^2}=\frac{a^2+ac}{ac+c^2}=\frac{a\left(a+c\right)}{c\left(a+c\right)}=\frac{a}{c}\left(dpcm\right)\)

\(2^{2\cdot25}=2^{50}\)   

\(3^{150}=\left(3^3\right)^{50}=27^{50}\)   

\(2^{50}< 27^{50}\)   

\(2^{2\cdot25}< 3^{150}\)

Ta có: \(x=-2y\)

\(\Leftrightarrow\frac{x}{-2}=\frac{y}{1}\)

Áp dụng t/c dãy tỉ số bằng nhau ta được:

\(\frac{x}{-2}=\frac{y}{1}=\frac{x+y}{-2+1}=\frac{10}{-1}=-10\)

\(\Rightarrow\hept{\begin{cases}x=20\\y=-10\end{cases}}\)

Cách khác : 

Thay x = -2y vào x + y = 10 ta có : \(-2y+y=10\Leftrightarrow-y=10\Leftrightarrow y=-10\)

Thay y = -10 ta có : \(x-10=10\Leftrightarrow x=20\)