Đại số lớp 7
Các câu hỏi tương tự
Tìm \(x\in Z\) biết :
\(a.\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x-1\right).\left(2x+1\right)}=\frac{49}{99}\)
\(b.\)
\(1-3+3^2-3^3+...+\left(-3^x\right)=\frac{1-9^{1006}}{4}\)
Thực hiện phép tính một cách hợp lí:
\(\frac{-2}{1.3}\)-\(\frac{2}{3.5}\)-\(\frac{2}{5.7}\)-\(\frac{2}{7.9}\)-\(\frac{2}{9.11}\)-\(\frac{2}{11.13}\)-\(\frac{2}{13.15}\)
Tính:
Afrac{7}{3}.frac{11}{16}+frac{10}{3}.frac{7}{16}-frac{7}{6}.frac{5}{8}
B1+2-3-4+5+6-7-8+.....+2005+2006-2007-2008+2009+2010
C(1-frac{1}{4})(1-frac{1}{9})(1-frac{1}{16})......(1-frac{1}{100000})
Dfrac{17frac{3}{4}.frac{17}{5}+3frac{2}{5}.82frac{1}{4}}{2.34-3.17}
Efrac{frac{2008}{2011}+frac{2009}{2010}+frac{2010}{2009}+frac{2011}{2008}+frac{2012}{503}}{frac{1}{2011}+frac{1}{2010}+frac{1}{2009}+frac{1}{2008}}
F(2-frac{2}{1.3})+(2-frac{2}{3.5})+(2-frac{2}{5.7})+.....+(2-frac{2}{2009.201...
Đọc tiếp
Tính:
A=\(\frac{7}{3}\).\(\frac{11}{16}\)+\(\frac{10}{3}\).\(\frac{7}{16}\)-\(\frac{7}{6}\).\(\frac{5}{8}\)
B=1+2-3-4+5+6-7-8+.....+2005+2006-2007-2008+2009+2010
C=(1-\(\frac{1}{4}\))(1-\(\frac{1}{9}\))(1-\(\frac{1}{16}\))......(1-\(\frac{1}{100000}\))
D=\(\frac{17\frac{3}{4}.\frac{17}{5}+3\frac{2}{5}.82\frac{1}{4}}{2.34-3.17}\)
E=\(\frac{\frac{2008}{2011}+\frac{2009}{2010}+\frac{2010}{2009}+\frac{2011}{2008}+\frac{2012}{503}}{\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}}\)
F=(2-\(\frac{2}{1.3}\))+(2-\(\frac{2}{3.5}\))+(2-\(\frac{2}{5.7}\))+.....+(2-\(\frac{2}{2009.2011}\))
Tính nhanh
\(A=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+\frac{1}{4.6}+...+\frac{1}{8.10}\)
Chứng minh M=\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}< 1\)
\(\frac{1}{2}\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)...........\left(1+\frac{1}{n\left(n+2\right)}\right)=\frac{2013}{2014}\)
Tìm n , n thuộc N
Tính Nhanh
\(\frac{\left(1+2+3+...+99+100\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right).\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)
Chứng mình rằng:
\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{99^2}\) < 1
CMR: \(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}< 1\)