\(A=\frac{1}{1.4}-\frac{1}{4.7}-\frac{1}{7.10}-...-\frac{1}{2011-2014}\)
\(A=\frac{1}{1.4}-\frac{1}{4.7}-\frac{1}{7.10}-.....-\frac{1}{2011.2014}\)
cố làm được nhé mọi người ahihi
Tìm x, biết:
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{125}{376}\)
Bài 1 : tính hợp lý :
a) A = \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}\)
b) B = \(\frac{1}{32}+\frac{1}{96}+\frac{1}{192}+\frac{1}{320}+\frac{1}{480}\)
Bài 1 : tính hợp lý :
a) A = \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}\)
b) B = \(\frac{1}{32}+\frac{1}{96}+\frac{1}{192}+\frac{1}{320}+\frac{1}{480}\)
a) \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}\frac{125}{376}\left(x€N\cdot\right)\)
b) \(\frac{3}{4}x-14\frac{2}{3}:\left(\frac{11}{15}+\frac{1111}{3535}+\frac{111111}{636363}\right)=12\)
\(Cho:S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n\left(n+3\right)}n\in Nsao\)
Chứng minh : S<1
M=\(\frac{1}{1.4}\)+\(\frac{1}{4.7}\)+\(\frac{1}{7.10}\)+...+\(\frac{1}{91.94}\)
So sánh A và B biết
a,A=6\(\left(x+\frac{1}{3}\right)^2\)
,B=-8-(3,75-x)2
b,A=\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\)
B=\(\left(\frac{1}{2}\right)^4\)
