\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)

\(=1-\frac{1}{20}=\frac{19}{20}\)

Vậy\(A=\frac{19}{20}\)

\(=\dfrac{1}{3}\left(1\times2\times3+2\times3\times3+...+19\times20\times3\right)\\ =\dfrac{1}{3}\left[1\times2\times\left(3-0\right)+2\times3\times\left(4-1\right)+...+19\times20\times\left(21-18\right)\right]\\ =\dfrac{1}{3}\left(1\times2\times3-1\times2\times3+2\times3\times4-...-18\times19\times20+19\times20\times21\right)\\ =\dfrac{1}{3}\times19\times20\times21=2660\)

Đặt \(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{18.19}+\frac{2}{19.20}\)

\(A=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)

\(A=2\left(1-\frac{1}{20}\right)\)

\(A=2.\frac{19}{20}=\frac{19}{10}\)

\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{18.19}+\frac{2}{19.20}\)

\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\right)\)

\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)

\(=2\left(1-\frac{1}{20}\right)\)

\(=2.\frac{19}{20}\)

\(=\frac{19}{10}\)

Ta có : \(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+......+\frac{2}{19.20}\)

\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{19.20}\right)\)

\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{19}-\frac{1}{20}\right)\)

\(=2\left(1-\frac{1}{20}\right)=2.\frac{19}{20}=\frac{19}{10}\)

Bài 1:

Đặt \(A=\frac{2}{1x2}+\frac{2}{2x3}+\frac{2}{3x4}+...+\frac{2}{18x19}+\frac{2}{19x20}\)

\(\frac{A}{2}=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{18x19}+\frac{1}{19x20}\)

\(\frac{A}{2}=\frac{2-1}{1x2}+\frac{3-2}{2x3}+\frac{4-3}{3x4}+...+\frac{19-18}{18x19}+\frac{20-19}{19x20}\)

\(\frac{A}{2}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}=1-\frac{1}{20}=\frac{19}{20}\)

\(A=\frac{2x19}{20}=\frac{19}{10}\)

Bài 2:

Đặt \(B=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{8x9}+\frac{1}{9x10}\)

Làm tương tự câu 1 có \(B=1-\frac{1}{10}=\frac{9}{10}\)

\(Bx100=\frac{9}{10}x100=90\)

=> \(\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]:\frac{1}{2}=1\)

=> \(\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]=\frac{1}{2}\)

=>  \(x+\frac{206}{100}=\frac{5}{2}:\frac{1}{2}=5\Rightarrow x=5-\frac{206}{100}=\frac{294}{100}=\frac{147}{50}\)

bài 1 đáp án là:19/10

2:147/50

đáp án bài 1 là: 19/10

đáp án bài 2 là 147/50

$\frac{1}{2\times 3}=\frac{3-2}{2\times 3}=\frac{3}{2\times 3}-\frac{2}{2\times 3}=\frac{1}{2}-\frac{1}{3}$

$\frac{1}{2\times 3}+\frac{1}{3\times 4}+\frac{1}{4\times 5}+...+\frac{1}{19\times 20}$

$=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}$$\frac{\mathrm{=}}{}$

Đáp án cuối cùng là 6 bạn nhé 

\(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+...+\dfrac{1}{18\times19}+\dfrac{1}{19\times20}\\ =\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{18}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{20}\\ =\dfrac{1}{2}-\dfrac{1}{20}=\dfrac{9}{20}\)

\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{18.19}+\frac{2}{19.20}\)

\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\right)\)

\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)

\(=2.\left(1-\frac{1}{20}\right)\)

\(=2.\frac{19}{20}=\frac{19}{10}\)

\(\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{19\cdot20}\)

\(=2\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\)

\(=2\cdot\left(1-\dfrac{1}{20}\right)\)

\(=2\cdot\dfrac{19}{20}\)

\(=\dfrac{19}{10}\)