1/ Tính tổng
a)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
b)\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
c)\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008+2010}\)
2/ Chứng tỏ rằng \(\frac{2n+1}{3n+2}\) và\(\frac{2n+3}{4n+8}\)là các phân số tối giản
3/ Cho \(A=\frac{n+2}{n-5}\)\(\left(n\in Z;n\ne5\right)\)Tìm n để \(A\in Z\)
4/ Chứng mình rằng:
a) \(\frac{a}{n\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\)\(\left(n,a\inℕ^∗\right)\)
b) Áp dụng câu a tính:
\(A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\) \(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)
\(C=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
5/ Với giá trị nào của \(x\in Z\)các phân số sau có giá trị là một số nguyên
a)\(A=\frac{3}{x-1}\) b)\(B=\frac{x-2}{x+3}\) c)\(C=\frac{2x+1}{x-3}\) d)\(D=\frac{x^2-1}{x+1}\)
a,\(\frac{2}{1.3}+...\frac{2}{99.101}\)
\(=\frac{3-1}{1.3}+...+\frac{101-99}{99.101}\)
\(=\frac{3}{1.3}-\frac{1}{1.3}+...+\frac{101}{99.101}-\frac{99}{99.101}\)
\(=\frac{1}{1}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{1}-\frac{1}{101}\)
\(\frac{100}{101}\)
Mình cần gấp, ai trả lời nhanh nhất mình k cho
\(3,\frac{n+2}{n-5}=\frac{n-5+7}{n-5}=\frac{n-5}{n-5}+\frac{7}{n-5}\)
=> Để phân số trên tối giản thì 7 phải chia hết cho n - 5
hay n - 5 là ước của 7
Từ đó ta suy ra được
n - 5 = 7 => n = 7+5= 12
n -5= -7 => n = -7 +5 = -2
n - 5 = 1 => n = 1 +5 =6
n - 5= -1 => n = -1 +5 = 4
Vậy để trên là phân số thì n phải bằng: 12 ; -2; 6;4
4. a)\(Tacó:\frac{1}{n}-\frac{1}{n+1}=\frac{1.\left(n+1\right)}{n.\left(n+1\right)}-\frac{1.n}{n.\left(n+1\right)}=\frac{n+1-n}{n.\left(n+1\right)}\\ \frac{1}{n.\left(n+1\right)}\)Mẫu số chung : n. (n+1)
Vậy:\(\frac{1}{n.\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
\(4,b:\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\\ \)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)
\(4,b\)\(B=\frac{5}{1.4}+\frac{5}{4.7}+...\frac{5}{100.103}\)
\(\frac{2}{5}B=\frac{3}{1.4}+...+\frac{3}{100.103}\)
\(\frac{2}{5}B=\frac{4-1}{1.4}+....+\frac{103-100}{100.103}\)
\(\frac{2}{5}B=\frac{1}{1}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{103}\)
\(\frac{2}{5}B=\frac{1}{1}-\frac{1}{103}\)
\(\frac{2}{5}B=\frac{102}{103}\)
\(B=\frac{102}{103}:\frac{2}{5}=\frac{255}{103}\)
\(A=\frac{2}{2.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(A=1-\frac{1}{101}\)
\(A=\frac{100}{101}\)
\(B=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
\(B=\frac{2}{1.3}\cdot\frac{5}{2}+\frac{2}{3.5}\cdot\frac{5}{2}+\frac{2}{5.7}\cdot\frac{5}{2}+...+\frac{2}{99.101}\cdot\frac{5}{2}\)
\(B=\frac{5}{2}\cdot\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(B=\frac{5}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(B=\frac{5}{2}\cdot\left(1-\frac{1}{101}\right)\)
\(B=\frac{5}{2}\cdot\frac{100}{101}\)
\(B=\frac{250}{101}\)
\(C=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(C=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(C=2\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(C=2\cdot\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(C=2\cdot\frac{502}{2005}\)
\(C=\frac{1004}{1005}\)
\(D=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(D=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(D=\frac{1}{2}-\frac{1}{100}\)
\(D=\frac{49}{100}\)
\(E=\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{101.103}\)
\(E=\frac{5}{3}\cdot\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{101.103}\right)\)
\(E=\frac{5}{2}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(E=\frac{5}{3}\cdot\left(1-\frac{1}{103}\right)\)
\(E=\frac{5}{3}\cdot\frac{102}{103}\)
\(E=\frac{170}{103}\)
\(F=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(F=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
4,b,\(C=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{249}\)
\(C=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(\frac{1}{2}C=\frac{2}{3.5}+...+\frac{2}{49.51}\)
\(\frac{1}{2}C=\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{49}-\frac{2}{51}\)
\(\frac{1}{2}C=\frac{2}{3}-\frac{2}{51}\\ \frac{1}{2}C=\frac{32}{51}\\ C=\frac{64}{51}\)
