Các câu hỏi tương tự
Rút gọn biểu thúc
\(\frac{A}{B}=\frac{\frac{1}{1\left(2n-1\right)}+\frac{1}{3\left(2n-3\right)}+\frac{1}{5\left(2n-5\right)}+...+\frac{1}{\left(2n-3\right).3}+\frac{1}{\left(2n-1\right).1}}{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2n-1}}\)
Rút gọn
\(\frac{A}{B}\)=\(\frac{\frac{1}{1\left(2n-1\right)}+\frac{1}{3\left(2n-3\right)}+\frac{1}{5\left(2n-5\right)}+...+\frac{1}{\left(2n-3\right)3}+\frac{1}{\left(2n-1\right)1}}{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2n-1}}\)
4 tháng 10 2017 lúc 20:50
1. Chứng minh : B left(1-frac{2}{6}right).left(1-frac{2}{12}right).left(1-frac{2}{20}right)...left(1-frac{2}{nleft(n+1right)}right)frac{1}{3}2. cho M frac{1}{1.left(2n-1right)}+frac{1}{3.left(2n-3right)}+frac{1}{5.left(2n-5right)}+...+frac{1}{left(2n-3right).3}+frac{1}{left(2n-1right).1}N 1+frac{1}{3}+frac{1}{5}+...+frac{1}{2n-1}Rút gọn frac{M}{N}
Đọc tiếp
1. Chứng minh : B = \(\left(1-\frac{2}{6}\right).\left(1-\frac{2}{12}\right).\left(1-\frac{2}{20}\right)...\left(1-\frac{2}{n\left(n+1\right)}\right)>\frac{1}{3}\)
2. cho M = \(\frac{1}{1.\left(2n-1\right)}+\frac{1}{3.\left(2n-3\right)}+\frac{1}{5.\left(2n-5\right)}+...+\frac{1}{\left(2n-3\right).3}+\frac{1}{\left(2n-1\right).1}\)
N = \(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2n-1}\)
Rút gọn \(\frac{M}{N}\)
A\(=\frac{1}{1.\left(2m-1\right)}+\frac{1}{3\left(2n-3\right)}+...+\frac{1}{\left(2n-3\right)3}+\frac{1}{\left(2n-1\right)1}\)
B\(=1+\frac{1}{3}+...+\frac{1}{2n-1}\)
tính A:B
Rút gọn:
B= \(\frac{1^2}{2^2-1}.\frac{3^2}{4^2-1}.\frac{5^2}{6^2-1}....\frac{\left(2n+1\right)^2}{\left(2n+2\right)^2-1}\)
Rút gọn biểu thức :
a) \(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{4}\right).\left(1+\frac{1}{16}\right)...\left(1+\frac{1}{2^{2n}}\right)\)
b) \(\left(10+1\right).\left(10^2+1\right)\left(10^3+1\right)...\left(10^{2n}+1\right)\)
Chứng minh: \(\frac{3}{\left(1x2\right)}+\frac{5}{\left(2x3\right)}+...+\frac{2n+1}{\left(n\left(n+1\right)\right)^2}=\frac{n\left(n+2\right)}{\left(n+1\right)^2}\)
Rút gọn \(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
a) Chứng minh rằng:\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
b) Áp dụng chứng minh rằng nếu \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\) thì \(\frac{1}{a^{2n+1}}+\frac{1}{b^{2n+1}}+\frac{1}{c^{2n+1}}=\frac{1}{a^{2n+1}+b^{2n+1}+c^{2n+1}}\) với mọi n thuộc N