Kết quả của phép tính \(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\) là :
(A) \(\dfrac{17}{60}\) (B) \(\dfrac{13}{60}\) (C) \(\dfrac{7}{60}\) (D) \(\dfrac{23}{60}\)
Hãy chọn kết quả đúng ?
a) Chứng tỏ rằng với \(n\in\mathbb{N},n\ne0\) thì :
\(\dfrac{1}{n\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\)
b) Áp dụng kết quả ở câu a) để tính nhanh :
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{9.10}\)
a) \(\forall\)n \(\in\) N* ta có :
\(\dfrac{1}{n\left(n+1\right)}=\dfrac{n+1-n}{n\left(n+1\right)}=\dfrac{n+1}{n\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\) (đpcm)
Trả lời bởi Mai Hà Chi
Tính nhanh :
\(A=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
A = 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56
A = (1/6 + 1/12) + 1/20 + 1/30 + 1/42 + 1/56
A = 1/4 + 1/20 + 1/30 + 1/42 + 1/56
A = (1/4 + 1/20) + 1/30 + 1/42 + 1/56
A = 3/10 + 1/30 + 1/42 + 1/56
A = (3/10 + 1/30) + 1/42 + 1/56
A = 1/3 + 1/42 + 1/56
A = (1/3 + 1/42) + 1/56
A = 5/14 + 1/56
A = 3/8
Trả lời bởi Bùi Khánh VânTính nhanh :
\(B=\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
\(B=\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
\(=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)
\(=\dfrac{1}{2}.\dfrac{10}{39}=\dfrac{5}{39}\)
Vậy \(B=\dfrac{5}{39}\)
Trả lời bởi Hàn Thất LụcTính nhanh :
\(C=\dfrac{1}{2}+\dfrac{1}{14}+\dfrac{1}{35}+\dfrac{1}{65}+\dfrac{1}{104}+\dfrac{1}{152}\)
\(C=\dfrac{1}{2}+\dfrac{1}{14}+\dfrac{1}{35}+\dfrac{1}{65}+\dfrac{1}{104}+\dfrac{1}{152}\)
\(=\dfrac{2}{4}+\dfrac{2}{28}+\dfrac{2}{70}+\dfrac{2}{130}+\dfrac{2}{208}+\dfrac{2}{304}\)
\(=\dfrac{2}{1.4}+\dfrac{2}{4.7}+\dfrac{2}{7.10}+\dfrac{2}{10.13}+\dfrac{2}{13.16}+\dfrac{2}{16.19}\)
\(=\dfrac{2}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{19}\right)\)
\(=\dfrac{2}{3}.\left(1-\dfrac{1}{19}\right)\)
\(=\dfrac{2}{3}.\dfrac{18}{19}=\dfrac{12}{19}\)
Trả lời bởi Hàn Thất Lục
Chứng tỏ rằng :
\(D=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{10^2}< 1\)
Ta có :
1/2^2<1/1.2
1/3^2<1/2.3
...
suy ra : 1/2^2+1/3^2+...+1/10^2<1/1.2+1/2.3+...+1/9.10
vì 1/1.2+1/2.3+...+1/9.10<1 nên 1/2^2+1/3^2+...+1/10^2<1
Trả lời bởi MonKey D. Luffy
(C) 7/60
Trả lời bởi hồ quỳnh anh