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D=1.2+2.3+3.4+...+19.20
=>3D=1.2.3+2.3.3+3.4.3+...+19.20
=1.2.3+2.3(4-1)+3.4(5-2)+...+19.20(21-18)
=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+19.20.21-18.19.20
=>3D=1.2.3+2.3.3+3.4.3+...+19.20
=1.2.3+2.3(4-1)+3.4(5-2)+...+19.20(21-18)
=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+19.20.21-18.19.20
=19.20.21=7980
=>D=7980:3=2660
Vậy D=2660
Đặt \(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{18.19}+\frac{2}{19.20}\)
\(\Rightarrow A=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(\Rightarrow A=2.\left(1-\frac{1}{20}\right)\)
\(\Rightarrow A=2.\frac{19}{20}\)
\(\Rightarrow A=\frac{19}{10}\)
2.(1/1.2+1/2.3+.....+1/18.19+1/19.20)
2.(1/1-1/2+1/2-1/3+......+1/19-1/20)
2.(1/1-1/20)= 2.19/20=19/10
\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{19\cdot20}\right)\div x=\frac{9}{10}\)
\(\Leftrightarrow\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\right)\div x=\frac{9}{10}\)
\(\Leftrightarrow\left(\frac{1}{1}-\frac{1}{20}\right)\div x=\frac{9}{10}\)
\(\Leftrightarrow\frac{19}{20}\div x=\frac{9}{10}\)
\(\Leftrightarrow x=\frac{19}{18}\)
Sửa đề : \(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}\right):x=\frac{9}{10}\)
\(\Leftrightarrow VT=\frac{9}{10}x\)
\(\Leftrightarrow\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\right)=\frac{9}{10}x\)
\(\Leftrightarrow\left(1-\frac{1}{20}\right)=\frac{9}{10}x\Leftrightarrow\frac{19}{20}=\frac{9}{10}x\)
\(\Leftrightarrow\frac{19}{20}=\frac{18x}{20}\) Khử mẫu ta đc : \(\Leftrightarrow18x=19\Leftrightarrow x=\frac{19}{18}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{19.20}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\)
\(=1-\frac{1}{20}=\frac{19}{20}\)
=1-1/2+1/2-1/3+1/3-1/4+.........+1/18-1/19+1/19-1/20
=1-1/20
=19/20
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}\)
\(=\frac{9}{20}\)
Ta có công thức :\(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}\)
\(=\frac{9}{20}\)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{19.20}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}\right)\)
\(=2.\left(1-\frac{1}{20}\right)\)
\(=2.\frac{19}{20}=\frac{19}{10}\)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{19.20}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{1}-\frac{1}{20}\)
\(=\frac{19}{20}\)
Lời giải:
$A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+....+\frac{19-18}{18.19}+\frac{20-19}{19.20}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}$
$=1-\frac{1}{20}=\frac{19}{20}$
19.27 + 47.81 + 19.20
= 19 .( 27 + 20) + 47.81
= 19 . 47 + 47 .81
= 47 .( 19+81)
= 47.100
= 4700
19.27 + 47.81 +19.20
=19.(27+20)+47.81
=19.47+47.81
=47.(19+81)
=47.100
4700