so sánh ;
A=\(\frac{1}{5^1}+\frac{1}{5^2}+....+\frac{1}{5^{2014}}+\frac{1}{5^{2015}}\)với \(\frac{1}{4}\)
Câu 1:Rút gọn các biểu thức:
A=\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{97.99}-\frac{5}{4}.\frac{13}{99}+\frac{5}{99}.\frac{1}{4}\)
Câu 2: So sánh:
A=\(\frac{2013}{2014}+\frac{2016}{2015}\)và \(\frac{2014}{2015}+\frac{2017}{2016}\)
Câu 3: Cho f(x)=ax2+bx+c. Biết 7a+b=0. Chứng minh rằng: f(10).f(-3)\(\ge\)0
Tính A= \(\left[\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\right]:\frac{2014}{2015}\)
So sánh 199110 với 99612
Bài 1 : Tính :
a)\(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right)\times230\frac{1}{5}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right)\div\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
b) \(\frac{2^{12}\times3^5-4^6\times9^2}{\left(2^4\times3\right)^6+8^4\times3^5}-\frac{5^{10}\times7^3-25^5\times49^2}{\left(125\times7\right)^3+5^9\times14^3}\)
c)P=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2016}}{\frac{2015}{1}+\frac{2014}{2}+\frac{2013}{3}+....+\frac{1}{2015}}\)
Cho A = 1+\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{4026}\) ; B = \(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{4025}\) . So sánh \(\frac{A}{B}với1\frac{2013}{2014}\)
Cho A = 1+\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{4026}\) ; B = \(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{4025}\) . So sánh \(\frac{A}{B}với1\frac{2013}{2014}\)
Cho \(A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{4026},B=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{4025}\) . So sánh \(\frac{A}{B}=1\frac{2013}{2014}\)
Cho \(A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{4026}\)
\(B=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{4025}\)
So sánh \(\frac{A}{B}\)với \(1\frac{2013}{2014}\)
\(A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{4026}\)
\(B=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{4025}\)
So sánh \(\frac{A}{B}\)với \(1\frac{2013}{2014}\)