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Xin lỗi máy tớ chỉ có cách viết phân số thế này / thông cảm
Ta có : A= 1/1 -1/2 + 1/2 -1/3 + 1/3 - 1/4 + 1/4 -1/5 +... + 1/19 - 1/20
=> A= 1/1 - 1/20
=> A = 19/20
Vậy A = 19/20
A= 1/1-1/2+1/2-1/3+1/4-1/5+...+1/101-1/102
A=1-1/102=102/102-1/102=101/102
ý b thì chờ mình tí tìm cách lập luận đã nhé
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}+\frac{1}{101.102}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{102}\)
\(A=1-\frac{1}{102}\)
\(A=\frac{101}{102}\)
\(B=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=\frac{1}{1}-\frac{1}{100}\)
\(B=\frac{99}{100}\)
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Bài 1
a) \(P=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)
b) \(S=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{33}{99}-\frac{1}{99}\)
\(=\frac{32}{99}\)
c)\(Q=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}\)
\(=\frac{10}{20}-\frac{1}{20}\)
\(=\frac{9}{20}\)
Tk mình nha!!
Câu 2:
\(P=\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{99}\right)\)
\(=\left(\frac{2}{2}+\frac{1}{2}\right).\left(\frac{3}{3}+\frac{1}{3}\right).\left(\frac{4}{4}+\frac{1}{4}\right)...\left(\frac{99}{99}+\frac{1}{99}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{100}{99}\)
\(=\frac{3\cdot4\cdot5...100}{2.3.4...99}\)
\(=\frac{3\cdot100}{2}\)
\(=\frac{300}{2}=150\)
...
= 1/2-1/3+1/3-1/4+...+ 1/19-1/20
= 1/2-1/20
=9/20
có phải như thế này ko bn
\(A=\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{19.20}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{19}-\frac{1}{20}=\frac{1}{2}-\frac{1}{20}\)
A = \(\frac{9}{20}\)
\(B=\frac{1}{99.100}-\frac{1}{98.99}-\frac{1}{97.98}-.....-\frac{1}{1.2}=-\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{99.100}\right)\)
\(B=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)=-\left(1-\frac{1}{100}\right)\)
B = \(-\frac{99}{100}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{99-98}{98.99}+\frac{100-99}{99.100}\)
\(=\frac{2}{1.2}-\frac{1}{1.2}+\frac{3}{2.3}-\frac{2}{2.3}+\frac{4}{3.4}-\frac{3}{3.4}+\frac{5}{4.5}-\frac{4}{4.5}+...+\frac{99}{98.99}-\frac{98}{98.99}+\frac{100}{99.100}-\frac{99}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(A=1-\frac{1}{6}=\frac{5}{6}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{n}-\frac{1}{n+1}\)
\(B=1-\frac{1}{n+1}=\frac{n}{n+1}\)
A = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\)
= \(1-\frac{1}{20}\)
= \(\frac{19}{20}\)
Vậy A = \(\frac{19}{20}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\)
\(A=\frac{1}{1}-\frac{1}{20}=\frac{19}{20}\)
Vậy A = 19/20