\(\dfrac{\left(-1\right)^4}{3}=\dfrac{1}{3}\) hoặc \(\dfrac{-1^4}{3}=\dfrac{1}{3}\)
\(\dfrac{\left(-1\right)^4}{3}=\dfrac{1}{3}\) hoặc \(\dfrac{-1^4}{3}=\dfrac{1}{3}\)
So sánh:
a, \(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{50}}\) và \(\dfrac{1}{2}\)
b, \(\dfrac{1}{4}-\dfrac{1}{4^2}+\dfrac{1}{4^3}-...+\dfrac{1}{4^{99}}\) và \(\dfrac{1}{12}\)
c, \(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{50}{3^{50}}\) và \(\dfrac{3}{4}\)
a ) (-\(\)\(\dfrac{4}{3}\)x + \(\dfrac{1}{2}\))2 = \(\dfrac{1}{4}\)
b) 3 x + 3 x+2= 20
c) ( \(\dfrac{1}{2}\)) x+ (\(\dfrac{1}{2}\))x+2= \(\dfrac{5}{4}\)
Giúp mk với !!! Ngày mai mk làm kiểm tra rồi !!! PLEASE
Tính:
a) S=1.2+2.3+3.4+...+99.100
b) B=\(\dfrac{49^{24}.125^{17}.2^8-5^{30}.7^{49}.4^5}{5^{29}.16^2.7^{48}}\)
c) C=\(\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}\right).3^5+\left(\dfrac{1}{3^5}+\dfrac{1}{3^6}+\dfrac{1}{3^7}+\dfrac{1}{3^8}\right).3^9+...+\left(\dfrac{1}{3^{97}}+\dfrac{1}{3^{98}}+\dfrac{1}{3^{99}}+\dfrac{1}{3^{100}}\right).3^{101}\)
d) D= \(3-3^2+3^3-3^4+...+3^{2017}-3^{2018}\)
a) \(x:\left(-\dfrac{1}{2}\right)^3=-\dfrac{1}{2}\)
b)\(\left(\dfrac{3}{4}\right)^2.x=\left(\dfrac{3}{4}\right)^5\)
ch/m:
\(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+\dfrac{4}{3^4}+...+\dfrac{100}{3^{100}}< \dfrac{3}{4}\)
cC N GIÚP MK NHÉ
Tìm \(n\in\) N*, biết: \(2n\div\left(1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+4+...+n}\right)=2020\)
chung mk:
\(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{100}{3^{100}}< \dfrac{3}{4}\)
bài 1 tính
a/ \(\left|1\dfrac{1}{2}-2\dfrac{1}{3}\right|^2+\left|1-\dfrac{5}{4}\right|\)
b/\(\left|\left(-1\right)^{100}-2\dfrac{1}{3}\right|+\dfrac{5}{6}\)
c/\(\left|\left(\dfrac{2}{3}\right)^{-1}-\dfrac{7}{4}\right|-\dfrac{7}{4}\)
d/ \(\left|x-5\right|+\left|x-8\right|\) với 5 \< x>/
K = \(\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+\dfrac{4}{5^4}+\dfrac{5}{5^5}+...+\dfrac{100}{5^{100}}\)
Mình đang cần gấp!!!
Tìm x biết:
1) \(\dfrac{x+1}{2}+\dfrac{x+1}{3}+\dfrac{x+1}{4}=\dfrac{x+1}{5}+\dfrac{x+1}{6}\)
2) \(\dfrac{x+1}{2009}+\dfrac{x+2}{2008}+\dfrac{x+3}{2007}=\dfrac{x+10}{2000}+\dfrac{x+11}{1999}+\dfrac{x+12}{1998}\)