Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A = \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2014^2}-1\right)\)
A = \(\left(-\frac{1.3}{2.2}\right)\left(-\frac{2.4}{3.3}\right)...\left(-\frac{2013.2015}{2014.2014}\right)\)
A = \(-\left[\frac{\left(1.2....2013\right)\left(3.4....2015\right)}{\left(2.3....2014\right)\left(2.3...2014\right)}\right]\)
A = \(-\left(\frac{2015}{2014.2}\right)\)
A = \(-\frac{2015}{4028}\)
C\(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{5.6}\)-\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)-\(\frac{1}{8.9}+\frac{1}{9.10}\)
c=\(\frac{1}{1}-\frac{1}{10}\)
c=\(\frac{9}{10}\)
còn a và b rễ lắm mình ko thích làm bài rễ đâu bạn cố chờ lời giải khác nhé!
b)
\(x-2.\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)
\(x-2\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(x-2=\frac{16}{9}:\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(x-2=8\)
=> x = 10
a)
\(A=\frac{1}{2}.\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{2013}{2014}\cdot\frac{2014}{2015}\cdot\frac{2015}{2016}\)
\(A=\frac{1}{2016}\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}\)
\(A=\frac{1.2.3...99}{2.3.4...100}\)
\(A=\frac{1}{100}\)
\(B=1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{72}\)
\(B=1+1+...+1+\left(\frac{1}{12}+\frac{1}{20}+...+\frac{1}{72}\right)\)
\(B=5.1+\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}\right)\)
\(B=5+\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{8}-\frac{1}{9}\right)\)
\(B=5+\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(B=5+\frac{2}{9}=\frac{47}{9}\)
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)......\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{99}{100}\)
\(=\frac{1.2.3.4....99}{2.3.4.5...100}\)
\(=\frac{1}{100}\)
a, \(A=\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(A=\frac{8}{9}-\left(\frac{1}{8}-\frac{1}{9}+\frac{1}{7}-\frac{1}{8}+\frac{1}{6}-\frac{1}{7}+...+1-\frac{1}{2}\right)\)
\(A=\frac{8}{9}-\left(1-\frac{1}{9}\right)\)
\(A=\frac{8}{9}-\frac{8}{9}\)
\(A=0\)
c) x=-2 nha
d) =\(\frac{1}{5.6}\)+\(\frac{1}{6.7}\)+......+\(\frac{1}{11.12}\)
=\(\frac{1}{5}\)-\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{7}\)+.....+\(\frac{1}{11}\)-\(\frac{1}{12}\)
=\(\frac{1}{5}\)-\(\frac{1}{12}\)= \(\frac{7}{60}\)
a/ \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=1-\frac{1}{10}\)
=> \(A=\frac{9}{10}\)
b/ \(A=\frac{n+2}{n-5}=\frac{n-5+7}{n-5}=\frac{n-5}{n-5}+\frac{7}{n-5}\)
=> \(A=1+\frac{7}{n-5}\)
Để A nguyên => 7 chia hết cho n-5 => n-5=(-7; -1; 1; 7)
=> n=(-2; 4, 6, 8)
a)
\(A=\left(\frac{1}{9}-\frac{1}{10}\right)-\left(\frac{1}{8}-\frac{1}{9}\right)-....-\left(1-\frac{1}{2}\right)=\frac{1}{9}-\frac{1}{10}-\frac{1}{8}+\frac{1}{9}-....-1+\frac{1}{2}\)
\(A=-\left(\frac{1}{10}+1\right)=-\frac{11}{10}\)
a)\(A=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\\ \Rightarrow A=-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}-\frac{1}{72}-\frac{1}{90}\\ \Rightarrow A=-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)Đặt \(B=\frac{1}{2}+\frac{1}{6}+...+\frac{1}{72}+\frac{1}{90}\)
\(\Rightarrow B=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)
\(\Rightarrow B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow B=1-\frac{1}{10}=\frac{9}{10}\)
Ta có : \(A=-B\)
\(\Rightarrow A=-\frac{9}{10}\)