\(A=1+1+\frac{1}{2!}+...+\frac{1}{2015!}\)
chịu ko bít làm típ
\(A=1+1+\frac{1}{2!}+...+\frac{1}{2015!}\)
chịu ko bít làm típ
Tính \(A=\sqrt{1+\left(1+\frac{1}{3}\right)^2}+\sqrt{1+\left(\frac{1}{2}+\frac{1}{4}\right)^2}+...+\sqrt{1=\left(\frac{1}{2016}+\frac{1}{2018}\right)^2}\)
Tính :
\(S=\frac{1}{2^2-1}.\frac{3^2}{4^2-1}.\frac{5^2}{6^2-1}.\frac{7^2}{8^2-1}...\frac{2015^2}{2016^2-1}\)
Cho \(A=\frac{1}{\left(x+y\right)^3}\left(\frac{1}{x^4}-\frac{1}{y^4}\right);B=\frac{1}{\left(x+y\right)^4}\left(\frac{1}{x^3}-\frac{1}{y^3}\right);C=\frac{1}{\left(x+y\right)^5}\left(\frac{1}{x^2}-\frac{1}{y^2}\right)\)
a) Rút gọn tổng A+B+C
b) Tính tổng A+B+C tại x=2016;y=2017
Cho f(x) = \(\frac{1}{2x-2x^2-1}\)
Tính giá trị biểu thức : \(f\left(\frac{1}{2016}\right)+f\left(\frac{2}{2016}\right)+f\left(\frac{3}{2016}\right)+...+f\left(\frac{2015}{2016}\right)+f\left(\frac{2016}{2016}\right)\)
1) (8x-5)(x2+2014)=0
2) \(\frac{2-x}{2015}-1=\frac{1-x}{2016}\)\(-\frac{x}{2017}\)
3) \(\frac{x-1}{2016}+\frac{x-2}{2015}=\frac{x-3}{2014}\)\(+\frac{x-4}{2013}\)
4) (2x-5)3-(3x-4)3+(x+1)3=0
Giải phương trình:
1,\(\left(x^2-x+1\right)^4+5x^4=6\left(x^2-x+1\right)^4\)
2,\(\frac{x+4}{x-1}+\frac{x-4}{x+1}=\frac{x-8}{x+2}+\frac{x+8}{x-2}+\frac{8}{3}\)
3,\(\left|x-2015\right|^{2015}+\left|x-2016\right|^{2016}=1\)
4,\(\frac{5}{2x-3}-\frac{1}{x+2}=\frac{5}{x-6}-\frac{7}{2x-1}\)
5,\(\left(x+2008\right)^4+\left(x+2009\right)^4=\frac{1}{8}\)
giải phương trình:
a)\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
b)\(\frac{x}{2012}+\frac{x+1}{2013}+\frac{x+2}{2014}+\frac{x+3}{2015}+\frac{x+4}{2016}=5\)
\(\frac{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)...\left(2016^4+\frac{1}{4}\right)}{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)...\left(2015^4+\frac{1}{4}\right)}\)
Tính: C=\(2\frac{1}{2016}\cdot\frac{1}{2015}-\frac{1}{672}\cdot3\frac{2014}{2015}-\frac{4}{2014\cdot2015}+\frac{4}{672}\)