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sau 3 phút có kết quả tùy bạn

A=\(\frac{37}{78}\)

B=\(\frac{9}{38}\)

a) A = \(\frac{5}{1.4}+\frac{29}{4.7}+\frac{71}{7.10}+....+\frac{10301}{100.103}\) (có 34 số hạng)

A = \(\frac{4+1}{1.4}+\frac{4.7+1}{4.7}+\frac{7.10+1}{7.10}+....+\frac{100.103+1}{103.100}\)

A = \(1+\frac{1}{1.4}+1+\frac{1}{4.7}+1+\frac{1}{7.10}+....+1+\frac{1}{100.103}\)

A = \(1.34+\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)\)

A = \(34+\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)

A = \(34+\frac{1}{3}.\left(1-\frac{1}{103}\right)\)

A = \(34+\frac{1}{3}\cdot\frac{102}{103}\)

A = \(34+\frac{34}{103}=\frac{3536}{103}\)

bạn làm hộ mik câu B với

\(Q=\frac{1.3}{3.5}+\frac{2.4}{5.7}+\frac{3.5}{7.9}+...+\frac{49.51}{99.101}\)

\(\Rightarrow Q=\frac{\left(2-1\right)\left(2+1\right)}{\left(2.2-1\right)\left(2.2+1\right)}+\frac{\left(3-1\right)\left(3+1\right)}{\left(3.2-1\right)\left(3.2+1\right)}+...+\frac{\left(50-1\right)\left(50+1\right)}{\left(50.2-1\right)\left(50.2+1\right)}\)

\(\Rightarrow Q=\frac{1}{4}-\frac{3}{8}\left(\frac{1}{2.2-1}-\frac{1}{2.2+1}\right)+\frac{1}{4}-\frac{3}{8}\left(\frac{1}{3.2-1}-\frac{1}{3.2+1}\right)+...+\)\(\frac{1}{4}-\frac{3}{8}\left(\frac{1}{50.2-1}-\frac{1}{50.2+1}\right)\)

\(\Rightarrow Q=49.\frac{1}{4}-49.\frac{3}{8}\left(\frac{1}{2.2-1}-\frac{1}{2.2+1}+...+\frac{1}{50.2-1}-\frac{1}{50.2+1}\right)\)

\(\Rightarrow Q=\frac{49}{4}-\frac{147}{8}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(\Rightarrow Q=\frac{49}{4}-\frac{147}{8}\left(\frac{1}{3}-\frac{1}{101}\right)\)

\(\Rightarrow Q=\frac{49}{4}-\frac{147}{8}.\frac{98}{303}\)

\(\Rightarrow Q=\frac{49}{4}-\frac{2401}{404}\)

\(\Rightarrow Q=\frac{637}{101}\)

1/5.6 + 1/6.7 + 1/7.8 +...+ 1/24.25

=1/5 - 1/6 + 1/6-1/7 +1/7-1/8 + ... + 1/24-1/25

=> Kết quả là: 1/5 - 1/25 = 4/25

b) 2/1.3 + 2/3.5 + 2/5.7 + 2/7.9+...+ 2/99.101

=2/1-2/3 + 2/3-2/5 + 2/5-2/7 + 2/7-2/9 + ... + 2/99-2/101

=> kết quả là 2/1 - 2/101 =200/101

a) \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}\)

=\(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)

=\(\frac{1}{5}-\frac{1}{25}\)

=\(\frac{4}{25}\)

b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)

=\(2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\right)\)

=\(2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)

=\(2.\left(\frac{1}{1}-\frac{1}{101}\right)\)

=\(2.\frac{100}{101}\)

=\(\frac{200}{101}\)

a)\(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)

\(=\frac{1}{5}-\frac{1}{25}\)

\(=\frac{5}{25}-\frac{1}{25}\)

\(=\frac{4}{25}\)

b) \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}\)

\(=\frac{101}{101}-\frac{1}{101}\)

\(=\frac{100}{101}\)

a, 1/1.2+1/1.3+...+1/99.100

= 1-1/2+1/2-1/3+1/3+...+1/99-1/100

=1-1/100

=99/100

A = 99/100

B = 2014/6051

C = 99

D = 9/38

a, \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)

=2.(\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\))

=\(2.\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)

=\(\frac{2}{2}.\left(1-\frac{1}{101}\right)\)

\(=\frac{100}{101}\)

b, \(\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)

=\(5.\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\right)\)

=\(5.\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)

=\(\frac{250}{101}\)

\(=\frac{5}{2}.\frac{100}{101}\)

a,$\frac{2}{1.3}$+$\frac{2}{3.5}$+$\frac{2}{5.7}$+....+$\frac{2}{99.101}$

=>\(\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\)

=>\(\frac{1}{1}-\frac{1}{101}\)

=>\(\frac{100}{101}\)

b,

$\frac{5}{1.3}$+$\frac{5}{3.5}$+$\frac{5}{5.7}$+....+$\frac{5}{99.101}$

a, \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)

=>\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\)

=>\(\frac{1}{1}-\frac{1}{101}\)

=>\(\frac{100}{101}\)

b,\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+....+\frac{5}{99.101}\)

=>\(\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{99.101}\right)\)

=>\(\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{101}\right)\)

=>\(\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)\)

=>\(\frac{5}{2}.\frac{100}{101}\)

=>\(\frac{250}{101}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(A=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{99}-\frac{1}{100}\right)\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=1-\frac{1}{100}\)

\(A=\frac{99}{100}\)

\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(B=2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)

\(B=2.\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(B=1.\left(1-\frac{1}{101}\right)\)

\(B=\frac{100}{101}\)

\(C=\frac{4}{4.7}+\frac{4}{7.10}+\frac{4}{10.13}+...+\frac{4}{73.76}\)

\(C=4.\left(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{73.76}\right)\)

\(C=4.\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\right)\)

\(C=\frac{4}{3}.\left(\frac{1}{4}-\frac{1}{76}\right)\)

\(C=\frac{4}{3}.\frac{9}{38}\)

\(C=\frac{6}{19}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\\ =\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+....+\left(\frac{1}{99}-\frac{1}{100}\right)\\ =1-\frac{1}{100}\\ =\frac{99}{100}\\ B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{99.101}\\ =\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\\ =\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)

\(C=\frac{4}{4.7}+\frac{4}{7.10}+....+\frac{4}{73.76}\\ =\frac{4}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{73.76}\right)\\ =\frac{4}{3}.\left(\frac{3}{4}-\frac{3}{76}\right)\\ =\frac{18}{19}\)

Học tốt Nghe!!

Chi tiết

\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}+\frac{2}{99.101}\\ \)

\(B=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{99-97}{97.99}+\frac{101-99}{99.101}\\ \)

\(B=\left(\frac{3}{1.3}-\frac{1}{3}\right)+\left(\frac{5}{3.5}-\frac{3}{3.5}\right)+\left(\frac{7}{5.7}-\frac{5}{5.7}\right)+...+\left(\frac{99}{97.99}-\frac{97}{97.99}\right)+\left(\frac{101}{99.101}-\frac{99}{99.101}\right)\\ \)

\(B=\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+...+\left(\frac{1}{97}-\frac{1}{99}\right)+\left(\frac{1}{99}-\frac{1}{101}\right)\\ \)

\(B=1+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{97}-\frac{1}{97}\right)+\left(\frac{1}{99}-\frac{1}{99}\right)-\frac{1}{101}\\ \)

\(B=1-\left(0\right)+\left(0\right)...+\left(0\right)-\frac{1}{101}=1-\frac{1}{101}=\frac{101-1}{101}=\frac{100}{101}\)

a cong tru loan nen ko hieu

b

A=5/1.4+5/4.7+..5/100.103

3/5.A=3/1.4+3/4.7+..+3/100.103

=1/1-1/4+1/4-1/7+...+1/100-1/103

=1-1/103=102/103

A=(5.102)/(3.103)=5.34/103

 nhung ma ko cothoi gian giai

\(S1=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\)

\(S1=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{101}=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)

\(S2=\frac{5}{1.3}+\frac{5}{3.5}+....+\frac{5}{99.101}\)

\(S2=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{101}\right)=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{5}{2}\cdot\frac{100}{101}=\frac{250}{101}\)

làm tắt thế ai mà hỉu đc