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\(A=\dfrac{1}{1.2}-\dfrac{1}{1.2.3}+\dfrac{1}{2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{3.4}-\dfrac{1}{3.4.5}+\dfrac{1}{99.100}-\dfrac{1}{99.100.101}\)
\(A=\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)-\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{99.100.101}\right)\)
\(A=\left(1-\dfrac{1}{100}\right)-\left(\dfrac{\dfrac{1}{1.2}-\dfrac{1}{100.101}}{2}\right)\)
Bấm máy nha
\(B=\dfrac{5}{1.2.3.4}+\dfrac{5}{2.3.4.5}+\dfrac{5}{3.4.5.6}+...+\dfrac{5}{98.99.100.101}\)
\(B=\dfrac{5}{3}.\left(\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+...+\dfrac{3}{98.99.100.101}\right)\)
\(B=\dfrac{5}{3}.\left(\dfrac{4-1}{1.2.3.4}+\dfrac{5-2}{2.3.4.5}+...+\dfrac{101-98}{98.99.100.101}\right)\)
\(B=\dfrac{5}{3}.\left(\dfrac{4}{1.2.3.4}-\dfrac{1}{1.2.3.4}+\dfrac{5}{2.3.4.5}-\dfrac{2}{2.3.4.5}+...+\dfrac{101}{98.99.100.101}-\dfrac{98}{98.99.100.101}\right)\)
\(B=\dfrac{5}{3}.\left(\dfrac{1}{1.2.3}-\dfrac{1}{99.100.101}\right)\)
\(B=\dfrac{5}{3}.\dfrac{166649}{999900}\approx0,3\)
1)Tính
a)\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+..........+\dfrac{1}{9.10}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
b)\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.........+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..............+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
2) tìm x
\(a\)) \(\dfrac{2}{5}+\dfrac{4}{5}x-\dfrac{7}{5}\)\(=\dfrac{9}{5}\)
\(\dfrac{4}{5}x+\dfrac{7}{5}=\dfrac{9}{5}-\dfrac{2}{5}\)
\(\dfrac{4}{5}x+\dfrac{7}{5}=\dfrac{7}{5}\)
\(\dfrac{4}{5}x=\dfrac{7}{5}-\dfrac{7}{5}\)
\(\dfrac{4}{5}x=0\)
\(x=0:\dfrac{4}{5}\)
\(x=0\)
b)\(\dfrac{2}{5}x-\dfrac{6}{4}=\dfrac{8}{5}\)
\(\dfrac{2}{5}x=\dfrac{8}{5}+\dfrac{6}{4}\)
\(\dfrac{2}{5}x=\dfrac{31}{10}\)
\(x=\dfrac{31}{10}:\dfrac{2}{5}\)
\(x=\dfrac{31}{4}\)
1. Tính:
a. \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
= \(\dfrac{1}{1}-\dfrac{1}{10}\)
= \(\dfrac{10}{10}-\dfrac{1}{10}\)
= \(\dfrac{9}{10}\)
b. \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
= \(\dfrac{1}{1}-\dfrac{1}{100}\)
= \(\dfrac{100}{100}-\dfrac{1}{100}\)
= \(\dfrac{99}{100}\)
2. Tìm x, biết:
a. \(\dfrac{2}{5}+\dfrac{4}{5}x-\dfrac{7}{5}=\dfrac{9}{5}\)
\(\dfrac{4}{5}x-\dfrac{7}{5}=\dfrac{9}{5}-\dfrac{2}{5}\)
\(\dfrac{4}{5}x-\dfrac{7}{5}=\dfrac{7}{5}\)
\(\dfrac{4}{5}x=\dfrac{7}{5}+\dfrac{7}{5}\)
\(\dfrac{4}{5}x=\dfrac{14}{5}\)
\(x=\dfrac{14}{5}:\dfrac{4}{5}\)
\(x=\dfrac{14}{5}.\dfrac{5}{4}\)
\(x=14.\dfrac{1}{4}\)
\(x=\dfrac{14}{4}\)
Vậy \(x=\dfrac{14}{4}\)
b. \(\dfrac{2}{5}x-\dfrac{6}{4}=\dfrac{8}{5}\)
\(\dfrac{2}{5}x=\dfrac{8}{5}+\dfrac{6}{4}\)
\(\dfrac{2}{5}x=\dfrac{32}{20}+\dfrac{30}{20}\)
\(\dfrac{2}{5}x=\dfrac{62}{20}\)
\(\dfrac{2}{5}x=\dfrac{31}{10}\)
\(x=\dfrac{31}{10}:\dfrac{2}{5}\)
\(x=\dfrac{31}{10}.\dfrac{5}{2}\)
\(x=\dfrac{31}{2}.\dfrac{2}{2}\)
\(x=\dfrac{31}{2}.1\)
\(x=\dfrac{31}{2}\)
Vậy \(x=\dfrac{31}{2}\)
bài này mk tự làm ko sao chép trên mạng
nếu thấy đúng thì tick đúng cho mk nha
\(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{99.100}\\ =2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\right)\\ =2.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =2.\left(\dfrac{1}{2}-\dfrac{1}{100}\right)=2.\dfrac{49}{100}=\dfrac{49}{50}\)
Đặt A = \(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{99.100}\)
\(\Rightarrow A=2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\right)\)
\(=2\left(\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+...+\dfrac{100-99}{99.100}\right)\)
\(=2\left(\dfrac{3}{2.3}-\dfrac{2}{2.3}+\dfrac{4}{3.4}-\dfrac{3}{3.4}+...+\dfrac{100}{99.100}-\dfrac{99}{99.100}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{100}\right)=1-\dfrac{1}{50}=\dfrac{49}{50}\)
\(a=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}\)
A = \(\dfrac{2}{1\times2}+\dfrac{2}{2\times3}+...+\dfrac{2}{98\times99}+\dfrac{2}{99\times100}\)
A = \(2\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{98\times99}+\dfrac{1}{99\times100}\right)\)
= \(2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
= \(2\left(1-\dfrac{1}{100}\right)\)
= \(2\times\dfrac{99}{100}=\dfrac{99}{50}\)
\(\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...............+\dfrac{2}{2008.2009}\)
\(=2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+................+\dfrac{1}{2008.2009}\right)\)
\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.................+\dfrac{1}{2008}-\dfrac{1}{2009}\right)\)
\(=2\left(1-\dfrac{1}{2009}\right)\)
\(=2.\dfrac{2008}{2009}=\dfrac{4016}{2009}\)
Giải:
a) \(\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^2}{3.4}...\dfrac{99^2}{99.100}\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{99}{100}\)
\(=\dfrac{1}{100}\)
Vậy giá trị của biểu thức trên là \(\dfrac{1}{100}\).
b) \(\left(\dfrac{2}{175}-\dfrac{7}{25}+\dfrac{3}{5}\right).\left(\dfrac{4}{11}+\dfrac{3}{121}-\dfrac{47}{121}\right)\)
\(=\left(\dfrac{2}{175}-\dfrac{7}{25}+\dfrac{3}{5}\right).\left(\dfrac{44}{121}+\dfrac{3}{121}-\dfrac{47}{121}\right)\)
\(=\left(\dfrac{2}{175}-\dfrac{7}{25}+\dfrac{3}{5}\right).\dfrac{0}{121}\)
\(=\left(\dfrac{2}{175}-\dfrac{7}{25}+\dfrac{3}{5}\right).0\)
\(=0\)
Vậy giá trị của biểu thức trên là 0.
c) \(-\dfrac{2}{5}\left(\dfrac{15}{17}-\dfrac{9}{15}\right)-\dfrac{2}{5}\left(\dfrac{2}{17}+\dfrac{-2}{5}\right)\)
\(=-\dfrac{2}{5}\left[\left(\dfrac{15}{17}-\dfrac{9}{15}\right)+\left(\dfrac{2}{17}+\dfrac{-2}{5}\right)\right]\)
\(=-\dfrac{2}{5}\left(\dfrac{15}{17}-\dfrac{9}{15}+\dfrac{2}{17}+\dfrac{-2}{5}\right)\)
\(=-\dfrac{2}{5}\left(1-1\right)\)
\(=-\dfrac{2}{5}.0\)
\(=0\)
Vậy giá trị của biểu thức trên là 0.
Chúc bạn học tốt!!!
\(\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^3}{3.4}...\dfrac{99^2}{99.100}\)
\(=\dfrac{1.1}{1.2}.\dfrac{2.2}{2.3}.\dfrac{3.3}{3.4}....\dfrac{99.99}{99.100}\)
\(=\dfrac{1.1.2.2.3.3.....99.99}{1.2.2.3.3.4....99.100}\)
\(=\dfrac{1.2.3...99}{1.2.3....99}.\dfrac{1.2.3....99}{2.3.4....100}=1.\dfrac{1}{100}=\dfrac{1}{100}\)
Lời giải:
Ta có:
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(A=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{100-99}{99.100}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
Vậy ta có đpcm.
\(P=\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{99\cdot100}\\ =2\cdot\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)\\ =2\cdot\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =2\cdot\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\\ =2\cdot\dfrac{99}{100}\\ =\dfrac{99}{50}\)
\(P=\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{99\cdot100}\\ =2\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)\\ =2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =2\left(1-\dfrac{1}{100}\right)=2\cdot\dfrac{99}{100}=\dfrac{99}{50}\)