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\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{9}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{8}{9}\)
\(A=\frac{1}{9}\)
\(\Rightarrow\)A= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}.\frac{7}{8}\frac{8}{9}\)
\(\Rightarrow\)A=\(\frac{1.2.3.4.5.6.7.8}{2.3.4.5.6.7.8.9}\)
\(\Rightarrow\)A=\(\frac{1}{9}\)
HỌC TỐT!!!
a) \(\left(x-\frac{2}{5}\right).\left(x+\frac{3}{7}\right)<0\)
\(\Rightarrow x-\frac{2}{5}<0\) hoặc \(x-\frac{2}{5}>0\)
\(x+\frac{3}{7}>0\) \(x+\frac{3}{7}<0\)
\(\Rightarrow x<\frac{2}{5}\) hoặc \(x>\frac{2}{5}\)
\(x>-\frac{3}{7}\) \(x<-\frac{3}{7}\)
\(\Rightarrow-\frac{3}{7} hoặc \(x\in rỗng\)
vậy \(-\frac{3}{7}
b) \(\frac{1}{2}-\left(\frac{1}{3}+\frac{1}{4}\right)\le x\le\frac{1}{24}-\left(\frac{1}{8}-\frac{1}{3}\right)\)
\(\frac{-1}{12}\le x\le\frac{1}{4}\)
\(\frac{-1}{12}\le x\le\frac{3}{12}\)
\(\Rightarrow x=\frac{-1}{12};0;\frac{1}{12};\frac{2}{12};\frac{3}{12}\)
A=\(\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).\left(\frac{1}{16}-1\right).............\left(\frac{1}{9801}-1\right).\left(\frac{1}{10000}-1\right)\)
A=\(\left(\frac{1-4}{4}\right).\left(\frac{1-9}{9}\right).\left(\frac{1-16}{16}\right).............\left(\frac{1-9801}{9801}\right).\left(\frac{1-10000}{10000}\right)\)
A=\(\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}.....................\frac{-9800}{9801}.\frac{-9999}{10000}\)
A=\(\frac{-1.3}{2^2}.\frac{-2.4}{3^2}.\frac{-3.5}{4^2}.....................\frac{-98.100}{99^2}.\frac{-99.101}{100^2}\)
A=\(\frac{\left[\left(-1\right).\left(-2\right).\left(-3\right)....................\left(-98\right).\left(-99\right)\right].\left(3.4.5............100.101\right)}{\left(2.3.4.........99.100\right).\left(2.3.4...............99.100\right)}\)
A=\(\frac{1.101}{100.2}\)=\(\frac{101}{200}\)
2
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.................+\frac{2}{x.\left(x+1\right)}=\frac{2015}{2017}\)
\(\frac{1}{3.2}+\frac{1}{6.2}+\frac{1}{10.2}+.................+\frac{2}{2.x.\left(x+1\right)}=\frac{1}{2}.\frac{2015}{2017}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.................+\frac{1}{x.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+..................+\frac{1}{x.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+..............+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{x+1}{2.\left(x+1\right)}-\frac{2}{2.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{\left(x+1\right)-2}{2.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{x-1}{2.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
=>\(\frac{x-1}{x+1}=\frac{2015}{2017}.\frac{1}{2}:\frac{1}{2}\)
\(\frac{x-1}{x+1}=\frac{2015}{2017}\)
=>x+1=2017
=>x=2018-1
=>x=2016
Vậy x=2016
Còn bài 3 em ko biết làm em ms lớp 6
Chúc anh học tốt
\(A = {1\over2}-{3\over4}+{5\over6}-{7\over12}={6\over12}-{9\over12}+{10\over12}-{7\over12}\)\(={0\over12}=0\)
Bài giải
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot\cdot\cdot\left(1-\frac{1}{2012}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2011}{2012}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot2011}{2\cdot3\cdot4\cdot...\cdot2012}\)
\(=\frac{1}{2012}\)
Bài giải
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot\cdot\cdot\left(1-\frac{1}{2012}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2011}{2012}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot2011}{2\cdot3\cdot4\cdot...\cdot2012}\) ( Sử dụng phương pháp khử )
\(=\frac{1}{2012}\)
\(a_{n-1}=\frac{1}{1+2+3+...+n}=\frac{2}{n\left(n+1\right)}\)=>\(1-a_{n-1}=1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
\(A=\left(1-\frac{2}{2.3}\right)\left(1-\frac{2}{3.4}\right)........\left(1-\frac{2}{2006.2007}\right)\)
\(=\left(\frac{1.4}{2.3}\right)\left(\frac{2.5}{3.4}\right)\left(\frac{3.6}{4.5}\right)........\left(\frac{2005.2008}{2006.2007}\right)\)\(=\frac{\left(1.2.3......2005\right)\left(4.5.6.....2008\right)}{\left(2.3.4.....2006\right)\left(3.4.5....2007\right)}=\frac{1.2008}{2006.3}=\frac{1004}{3009}\)
Nguyễn thị Thùy Dương làm đúng đấy