\(A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2015.2017}\)

\(A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)

\(A=1-\dfrac{1}{2017}=\dfrac{2016}{2017}\)

\(B=\dfrac{3}{1.4}+\dfrac{3}{5.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\)

\(B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\)

\(B=1-\dfrac{1}{103}=\dfrac{102}{103}\)

\(C=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{62.65}\)

\(3C=3\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{62.65}\right)\)

\(3C=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{62.65}\)

\(3C=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{62}-\dfrac{1}{65}\)

\(3C=\dfrac{1}{2}-\dfrac{1}{65}\)

\(3C=\dfrac{63}{130}\)

\(C=\dfrac{63}{130}:3=\dfrac{21}{130}\)

\(A=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{n\left(n+1\right)}\)

\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{n\left(n+1\right)}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{n}-\dfrac{1}{n+1}\)

\(=1-\dfrac{1}{n+1}\)

\(=\dfrac{n}{n+1}\)

Vì góc xOz < góc xOy (42 độ < 84 độ) nên tia Oz nằm giữa 2 tia Oy và Ox

=> góc xOz + góc yOz = góc xOy

42 độ + góc yOz = 84 độ

góc yOz = 84 độ - 42 độ

góc yOz = 42 độ

Ta có: góc yOz + góc yOz' = 180 độ (kề bù)

42 độ + góc yOz' = 180 độ

góc yOz' = 180 độ - 42 độ

góc yOz' = 138 độ

Vậy góc yOz' = 138 độ

Dụng cụ: thước dây,đồng hồ

B1: dùng thước dây để đo đường đi từ nhà đến trường,đồng hồ để đo thời gian từ nhà đến trường

B2: lấy quãng đường đo được chia cho thời gian đi hết quãng đường đó

B3: Báo cáo kết quả

Bài 2:

Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};....;\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

\(\Rightarrow A< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}=1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=2-\dfrac{1}{100}< 2\)

Vậy A < 2

Bài 3:

D = \(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right)....\left(1-\dfrac{1}{2015}\right)\)

\(=\dfrac{1}{2}.\dfrac{2}{3}......\dfrac{2014}{2015}\)

\(=\dfrac{1.2......2014}{2.3......2015}=\dfrac{1}{2015}\)

Bài 4:

A = \(\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}......\dfrac{899}{900}\)

\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}........\dfrac{29.31}{30.30}\)

\(=\dfrac{1.2.3......29}{2.3.4.......30}.\dfrac{3.4.5......31}{2.3.4.....30}\)

\(=\dfrac{1}{30}.\dfrac{31}{2}=\dfrac{31}{60}\)

a, \(\dfrac{2483-13}{4966-26}=\dfrac{2483-13}{2\left(2483-13\right)}=\dfrac{1}{2}\)

b, \(\dfrac{2727-101}{7575+303}=\dfrac{2525+202-101}{2525.3+303}=\dfrac{2525+101}{25.101.3+303}\)

\(=\dfrac{25.101+101}{25.303+303}=\dfrac{101\left(25+1\right)}{303\left(25+1\right)}=\dfrac{101}{303}=\dfrac{1}{3}\)

c, \(\dfrac{2002}{2000.16-1970}\)

\(=\dfrac{2002}{2000.15+2000-1970}\)

\(=\dfrac{2002}{2000.15+30}\)

\(=\dfrac{2002}{2000.15+2.15}\)

\(=\dfrac{2002}{15\left(2000+2\right)}=\dfrac{2002}{15.2002}=\dfrac{1}{15}\)

d,\(\dfrac{1.2.3+2.4.6+4.8.12+7.14.21}{1.3.6+2.6.12+4.12.24+7.21.42}\)

\(=\dfrac{1.2.3+2.4.6+4.8.12+7.14.21}{1.3.2.3+2.6.4.3+4.12.8.3+7.21.14.3}\)

\(=\dfrac{1.2.3+2.4.6+4.8.12+7.14.21}{3\left(1.2.3+2.4.6+4.8.12+7.14.21\right)}\)

\(=\dfrac{1}{3}\)

BCNN(2,3,15,3) = 30

\(\dfrac{1}{2}=\dfrac{1.15}{2.15}=\dfrac{15}{30};\dfrac{1}{3}=\dfrac{1.10}{3.10}=\dfrac{10}{30};\dfrac{1}{15}=\dfrac{1.2}{15.2}=\dfrac{2}{30};\dfrac{1}{3}=\dfrac{1.10}{3.10}=\dfrac{10}{30}\)

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