Tính:
a) \(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)
b) \(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
chứng minh rằng :
a) \(\frac{a}{n\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\) ( n , a ϵ N* )
b) áp dụng câu a tính ;
\(A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)
\(C=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
a) \(\frac{1}{n}-\frac{1}{n+a}=\frac{\left(n+a\right)-n}{n\left(n+a\right)}=\frac{a}{a\left(n+a\right)}\) (đpcm)
b) \(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
\(B=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{5}{3}.\left(1-\frac{1}{103}\right)=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)
\(C=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}=\frac{1}{3}-\frac{1}{51}=\frac{16}{51}\)
a. Chứng minh rằng
\(\frac{a}{n\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\left(n,a\in Nsao\right)\)
b. Áp dụng câu a tính:
A= \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
B= \(\frac{5}{1.4}+\frac{5}{4.7}+..+\frac{5}{100.103}\)
C= \(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
b) A=1/2.3+1/3.4+....+1/99.100
=> A=1/2-1/3+1/3-1/4+....+1/99-1/100
=> A=1/2-1/100
=> A=50/100-1/100
=> A=49/100
chứng minh rằng
\(\frac{a}{n\left(n+a\right)}\)=\(\frac{1}{n}\)-\(\frac{1}{n+a}\)
áp dụng tính
A=\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+.....+\(\frac{1}{99.100}\)
B=\(\frac{5}{1.4}\)+\(\frac{5}{4.7}\)+....+\(\frac{5}{100.103}\)
C=\(\frac{1}{15}\)+\(\frac{1}{35}\)+.....+\(\frac{1}{2499}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{2}-\frac{1}{100}\)
\(A=\frac{49}{100}\)
\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)
\(B=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
\(B=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(B=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{103}\right)\)
\(B=\frac{510}{103}\)
\(C=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(C=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)\)
\(C=\frac{8}{51}\)
\(\frac{5}{1.4}+\frac{5}{4.7}+............+\frac{5}{100.103}\)
\(\frac{5}{1.4}+\frac{5}{4.7}+................+\frac{5}{100.103}\)
\(\frac{1}{3}.\left(5-\frac{5}{4}+\frac{5}{4}-\frac{5}{7}+..............+\frac{5}{100}-\frac{5}{103}\right)\)
\(\frac{1}{3}.\left(5-\frac{5}{103}\right)\)
\(\frac{1}{3}.\left(\frac{510}{103}\right)=\frac{170}{103}\)
\(=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{103}\right)=\frac{5}{3}.\frac{102}{103}=\frac{....}{....}\)
xin lỗi mình ko có máy tính nên ko tính đc phép tính cuối
\(\frac{5}{1.4}+\frac{5}{4.7}+.....+\frac{5}{100.103}\)
=5/3.(1/1-1/4+1/4-1/7+...+1/100-1/103)
=5/3.(1/1-1/103)
=5/3.102/103
=170/103
=5/3.(1/1-1/4+1/4-1/7+...+1/100-1/103)
=5/3.(1/1-1/103)
=5/3.102/103
=170/103
\(S=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}=?\)
S = \(5-\frac{5}{4}+\frac{5}{4}-\frac{5}{7}+.......+\frac{5}{100}-\frac{5}{103}\)
S = \(5-\frac{5}{103}\)
S = \(\frac{510}{103}\)
tính nhanh:
\(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+.........+\frac{5}{100.103}\)
\(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{100.103}\)
\(=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(=\frac{5}{3}\left(1-\frac{1}{103}\right)\)
\(=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)
\(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{100.103}=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{103}\right)=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)
=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
Bài 3 Tính giá trị biểu thức\(\left(1_{ },5\right).\frac{-2}{3}+\left(2,5-\frac{3}{4}\right):1\frac{3}{4}\)
B=\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{100.103}\)
\(B=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
\(B=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(B=\frac{1}{3}.\left(1-\frac{1}{103}\right)\)
\(B=\frac{1}{3}.\frac{102}{103}\)
\(B=\frac{34}{103}\)
Bài 3: đổi ra phân số rồi tính, đổi:\(1,5=\frac{15}{10};2,5=\frac{25}{10};1\frac{3}{4}=\frac{7}{12}\)(cái này ko giải dùm, đổi ra như thek rồi tính nha)
\(B=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{100.103}\)
\(=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)\)
\(=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(=\frac{1}{3}.\left(1-\frac{1}{103}\right)\)
\(=\frac{1}{3}.\frac{102}{103}\)
\(=\frac{1}{1}.\frac{34}{103}=\frac{34}{103}\)
Tính:
\(A=\frac{5}{1.4}+\frac{29}{4.7}+\frac{71}{7.10}+...+\frac{10301}{100.103}\)