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\(\frac{1}{13}+\frac{3}{13\cdot23}+\frac{3}{23\cdot33}+...+\frac{3}{1993\cdot2003}\)
\(=\frac{1}{13}+\left[\frac{3}{13\cdot23}+\frac{3}{23\cdot33}+...+\frac{3}{1993\cdot2003}\right]\)
\(=\frac{1}{13}+\left[\frac{3}{10}\left[\frac{1}{13\cdot23}+\frac{1}{23\cdot33}+...+\frac{1}{1993\cdot2003}\right]\right]\)
\(=\frac{1}{13}+\left[\frac{3}{10}\left[\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+...+\frac{1}{1993}-\frac{1}{2003}\right]\right]\)
\(=\frac{1}{13}+\left[\frac{3}{10}\left[\frac{1}{13}-\frac{1}{2003}\right]\right]\)
\(=\frac{1}{13}+\left[\frac{3}{10}\cdot\frac{1990}{26039}\right]\)
\(=\frac{1}{13}+\frac{597}{26039}\)
\(=\frac{200}{2003}\)
Đặt A= 1/13 + 3/13.23 + 3/ 23.33 + ... + 3/1993.2003
A- 1/13 = 3/13.23 + 3/ 23.33 + ... + 3/1993.2003
10/3 ( A-1/3) = 10/3. (3/13.23 + 3/ 23.33 + ... + 3/1993.2003)
10/3A - 10/9 = 10/13.23 + 10/ 23.33 + ... + 10/1993.2003
10/3A - 10/9 = 1/13 - 1/23 + 1/23 - 1/33 +...+ 1/1993- 1/2003
10/3A = 1/13 - 1/2003 + 10/9
10/3 A= ?
đến đây bn tự làm nha
10/3A - 10/9 = 1/13
\(S=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)
S = 1/2 . ( 1/2 -1/2 + 1/6 -1/2 + ...+ 1/99 - 1/100)
S= 1/2 . (1-2 - 1/100)
S=1/2 . 49/100
S= 49/200
\(S=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{98.100}\)
=>2S=\(2.\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{98.100}\right)\)
=\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{98.100}\)
=\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\)
=\(\frac{1}{2}-\frac{1}{100}=\frac{50}{100}-\frac{1}{100}=\frac{49}{100}\)
=>S=\(\frac{49}{100}:2=\frac{49}{100}.\frac{1}{2}=\frac{49}{200}\)
Bài này khá ez thôi:
a) bạn sửa lại đề rồi làm theo cách làm của b,c,d nhé
b) Ta có: \(\left|x+1,1\right|+\left|x+1,2\right|+\left|x+1,3\right|+\left|x+1,4\right|\ge0\left(\forall x\right)\)
\(\Rightarrow5x\ge0\Rightarrow x\ge0\) khi đó:
\(PT\Leftrightarrow x+1,1+x+1,2+x+1,3+x+1,4=5x\)
\(\Leftrightarrow x=5\)
c,d tương tự nhé
c,\(\left|x+\frac{1}{1.3}\right|+\left|x+\frac{1}{3.5}+\right|+...+\left|x+\frac{1}{97.99}\right|\ge0\forall x\)
\(\Rightarrow50x\ge0\Rightarrow x\ge0\)Khi đó:
\(x+\frac{1}{1.3}+x+\frac{1}{3.5}+...+x+\frac{1}{97.99}=50x\)
\(\Rightarrow49x+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)=50x\)
\(\Leftrightarrow x=\frac{1}{2}\left(1-\frac{1}{99}\right)=\frac{49}{99}\)
\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{48.50}.\)
\(=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}....+\frac{2}{48.50}\right)\)
\(=\frac{1}{2}.\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{50-48}{48.50}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+.....+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{1}{2}.\frac{12}{25}=\frac{6}{25}\)
\(B=\frac{3}{1.4}+\frac{3}{4.7}+....+\frac{3}{97.100}\)
\(=\frac{4-1}{1.4}+\frac{7-4}{4.7}+....+\frac{100-97}{97.100}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(C=\frac{8}{7.14}+\frac{8}{14.21}+....+\frac{8}{91.98}\)
\(=\frac{7}{8}.\left(\frac{7}{7.14}+\frac{7}{14.21}+...+\frac{7}{91.98}\right)\)
\(=\frac{7}{8}.\left(\frac{1}{7}-\frac{1}{14}+\frac{1}{14}-\frac{1}{21}+.....+\frac{1}{91}-\frac{1}{98}\right)\)
\(=\frac{7}{8}.\left(\frac{1}{7}-\frac{1}{98}\right)\)
\(=\frac{7}{8}.\frac{13}{98}=\frac{13}{112}\)
a, \(\frac{1}{x}=\frac{1}{6}+\frac{y}{3}\)
\(\Rightarrow\frac{1}{x}=\frac{1}{6}+\frac{2y}{6}=\frac{1+2y}{6}\)
\(\Rightarrow1\cdot6=x\cdot\left(1+2y\right)\)
\(\Rightarrow x\left(1+2y\right)=6\)
\(\Rightarrow x;1+2y\inƯ\left(6\right)=\left\{-1;1;-2;2;-3;3;-6;6\right\}\)
ta có bảng :
| x | -1 | 1 | -2 | 2 | -3 | 3 | -6 | 6 |
| 1+2y | -6 | 6 | -3 | 3 | -2 | 2 | -1 | 1 |
| y | loại | loại | 2 | -1 | loại | loại | 1 | 0 |
vậy_
phần b tương tự
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=\frac{1}{2}\cdot\frac{98}{99}-\frac{1}{2}\cdot\frac{49}{100}\)
\(=\frac{1}{2}\left(\frac{98}{99}-\frac{49}{100}\right)=\frac{1}{2}\cdot\frac{4949}{9900}=\frac{4949}{19800}\)