Tính nhanh: \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
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tính nhanh
\(100-\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
tính nhanh: \(\frac{100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}}\)
Tách 100 thành 100 số 1
Ta có: TS=\(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)=100-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{100}=\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{100}\right)\)
=\(0+\frac{1}{2}+\frac{2}{3}+..+\frac{99}{100}=\frac{1}{2}+\frac{2}{3}+..+\frac{99}{100}\)=MS
=> Phân số trên=1
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tính nhanh \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+....+\frac{99}{100}}\)
tính nhanh
A= \(\frac{3}{1}+\frac{3}{1+2}+\frac{3}{1+2+3}+\frac{3}{1+2+3+4}+......+\frac{3}{1+2+3+4+.....+100}\)
bạn giải giúp mk bài này nhé
cầu xin bạn tại mk đang cần gấp huhuhu
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1) Tính nhanh \(\frac{\left(1+2+3+...+100\right).\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.,..................+\frac{1}{100}}\)
Tính nhanh, tính bằng cách hợp lí:
S=\(3+\frac{3}{1+2}+\frac{3}{1+2+3}+\frac{3}{1+2+3+4}+....+\frac{3}{1+2+3+...+100}\)
S=\(3\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+100}\right)\)
\(S=3\left(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{5050}\right)\)
\(S=3.\frac{1}{2}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{10100}\right)\)
\(S=\frac{3}{2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}\right)\)
\(S=\frac{3}{2}\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(S=\frac{3}{2}\left(1-\frac{1}{101}\right)\)
\(S=\frac{3}{2}.\frac{100}{101}=\frac{150}{101}\)
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Tính nhanh:\(\frac{\frac{1}{2}}{1+2}+\frac{\frac{1}{2}}{1+2+3}+\frac{\frac{1}{2}}{1+2+3+4}+...+\frac{\frac{1}{2}}{1+2+3+4+...+100}\)
Đặt A = \(\frac{\frac{1}{2}}{1+2}+\frac{\frac{1}{2}}{1+2+3}+...+\frac{\frac{1}{2}}{1+2+3+....+100}\)
= \(\frac{1}{2}\left(\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+...+\frac{1}{100.101:2}\right)\)
= \(\frac{1}{2}\left(\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{100.101}\right)\)
= \(\frac{1}{2}.2\left(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{100.101}\right)\)
= 1\(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{100}-\frac{1}{101}\right)\)
= \(\frac{1}{2}-\frac{1}{101}=\frac{101}{202}-\frac{2}{202}=\frac{99}{202}\)
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Tính nhanh:
(1+2+3+...+100) (\(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\)) (6,3.12-21.36,6)
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\)
Theo đề ta có: \(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}\right)\left(6,3.12-21.3,6\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)
\(=\frac{\left(1+2+3...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).0}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)
= 0
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Tính nhanh \(\frac{\left(1+2+3+.......+100\right).\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+................+\frac{1}{100}}\)
bài này mik giải rùi! ở câu hỏi tương tụ đó
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Tính nhanh:
1 + (\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\)) - \(\frac{100}{3^{100}}\)