Tính nhanh:
\(B=\dfrac{5}{1.4}+\dfrac{5}{4.7}+.....+\dfrac{5}{100.103}\)
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\(B=\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{1}{100.103}\)
\(B=\dfrac{5}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)=\dfrac{5}{3}\left(\dfrac{102}{103}\right)=\dfrac{170}{103}\)
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B= \(\dfrac{2}{1.4}+\dfrac{2}{4.7}+\dfrac{2}{7.9}+...+\dfrac{2}{100.103}\)
\(\dfrac{3}{2}\)B= \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\)
\(\dfrac{3}{2}\)B= \(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\) \(\dfrac{3}{2}\)B= \(\dfrac{102}{103}\) \(\)B= \(\dfrac{102}{103}:\dfrac{3}{2}\) B=\(\dfrac{68}{103}\)
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tính tổng: A dfrac{2}{1.3}+dfrac{2}{3.5}+dfrac{2}{5.7}+...+dfrac{2}{99.101} B dfrac{5}{1.3}+dfrac{5}{3.5}+dfrac{5}{3.7}+...+dfrac{5}{99.101}
C dfrac{1}{2.3}+dfrac{1}{3.4}+...+dfrac{1}{99.100} D dfrac{5}{1.4}+dfrac{5}{4.7}+...+dfrac{5}{100.103} E dfrac{1}{15}+dfrac{1}{35}+...+dfrac{1}{2499}
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tính tổng: A= \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\) B= \(\dfrac{5}{1.3}+\dfrac{5}{3.5}+\dfrac{5}{3.7}+...+\dfrac{5}{99.101}\)
C= \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\) D= \(\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{100.103}\) E= \(\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\)
A=2.(1/1.3 + 1/3.5 + 1/5.7 +.......+1/99.101)
=2.(1/1 + 1/3 + 1/5 + 1/5 + 1/7 +...+1/99 + 1/101)
=2.(1-1/101)
=2.(101/101-1/101)
=2.100/101
200/101
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B=2.(1/1.3+1/3.5+1/3.1+....+1/99.101)
=2.(1/1+1/3+1/3+1/5+1/3+1/7+....+1/99+1/101)
=2.(1/1+1/101)
=2.(101/101+1/101)
=2.102/101
=204/101
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C=1/2+1/3+1/3+1/4+....+1/99+1/100
=1/2+1/100
=50/100+1/100
=51/100
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\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}\)
\(\dfrac{1}{3}\)x(\(\dfrac{3}{1+4}\)+\(\dfrac{3}{4+7}\)+........+\(\dfrac{3}{101+103}\))
\(\dfrac{1}{3}\)x(\(\dfrac{1}{1}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{7}\)+.........+\(\dfrac{ }{ }\)\(\dfrac{1}{101}\)-\(\dfrac{1}{103}\))
\(\dfrac{1}{3}\)x(\(\dfrac{1}{1}\)-\(\dfrac{1}{103}\))
\(\dfrac{1}{3}\)x\(\dfrac{102}{103}\)
\(\dfrac{34}{103}\)
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\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}\)
\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\)
\(=\dfrac{1}{1}-\dfrac{1}{103}\)
\(=\dfrac{102}{103}\)
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\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{103}\right)=\dfrac{1}{3}.\dfrac{102}{103}=\dfrac{102}{309}=\dfrac{34}{103}\)
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6. Tính
\(A=\dfrac{4}{1.4}+\dfrac{4}{4.7}+\dfrac{4}{7.10}+...+\dfrac{4}{31.34}\)
\(B=1-5+5^2-5^3+5^4-...-5^{39}\)
a) Ta có: \(A=\dfrac{4}{1\cdot4}+\dfrac{4}{4\cdot7}+\dfrac{4}{7\cdot10}+...+\dfrac{4}{31\cdot34}\)
\(=\dfrac{4}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{31\cdot34}\right)\)
\(=\dfrac{4}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{31}-\dfrac{1}{34}\right)\)
\(=\dfrac{4}{3}\left(1-\dfrac{1}{34}\right)\)
\(=\dfrac{4}{3}\cdot\dfrac{33}{34}=\dfrac{22}{17}\)
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Tính nhanh: S1 = \(\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{97.100}\)
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`S_1 = 5/(1.4) + 5/(4.7) +...+ 5/(97.100)`
`S_1 = 5 (1/(1.4) + 1/(4.7) +...+ 1/(97.100))`
`S_1 = 5/3 (3/(1.4) + 3/(4.7) +...+ 3/(97.100))`
`S_1 = 5/3 (1 - 1/4 + 1/4 - 1/7 + ...+ 1/97 - 1/100)`
`S_1 = 5/3 (1 - 1/100)`
`S_1 = 5/3 . 99/100`
`S_1 = 33/20`
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Bài 5: Thực hiện phép tính \(\dfrac{2}{1.4}+\dfrac{2}{4.7}+...+\dfrac{2}{97.100}\)
=\(2.\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+.....+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
=\(2.\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\)
= \(2.\dfrac{99}{100}\)
=\(\dfrac{99}{50}\)
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Dạng toán nâng cao
Câu 1:Chứng minh rằngdfrac{a}{nleft(n+aright)}dfrac{1}{n}-dfrac{1}{n+a} ( n, a in N*)
Câu 2: Áp dụng tính:
Adfrac{1}{2.3}+dfrac{1}{3.4}+...+dfrac{1}{99.100}
Bdfrac{5}{1.4}+dfrac{5}{4.7}+...+dfrac{5}{100.103}
Cdfrac{1}{15}+dfrac{1}{35}+...+dfrac{1}{2499}
-Ai làm nhanh thì mình tick cho nha! Cảm ơn nhiều-
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Dạng toán nâng cao
Câu 1:Chứng minh rằng\(\dfrac{a}{n\left(n+a\right)}=\dfrac{1}{n}-\dfrac{1}{n+a}\) ( n, a \(\in\) N*)
Câu 2: Áp dụng tính:
\(A=\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(B=\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{100.103}\)
\(C=\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\)
-Ai làm nhanh thì mình tick cho nha! Cảm ơn nhiều-
Câu 1 :
1/n - 1/n + a = a + n/a ( a + n ) = a + n - a/a ( n + a ) = n/a ( a + n )
Câu 2 :
A = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +.......+ 1/99 - 1/100
= 1/1 - 1/100 = 99/100
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tính nhanh:
\(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+.........+\frac{5}{100.103}\)
\(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{100.103}\)
\(=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(=\frac{5}{3}\left(1-\frac{1}{103}\right)\)
\(=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)
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\(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{100.103}=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{103}\right)=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)
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=5/3.(3/1.4+3/4.7+3/7.10+...+3/100.103)
=5/3.(1-1/4+1/4-1/7+1/7-1/10+...+1/100-1/103)
=5/3.(1-1/103)=5/3.102/103=170/103
đáp số : 170/103
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