\(\left(\dfrac{1}{3}\right)^2\cdot\dfrac{1}{3\cdot9^2}=\dfrac{1}{9\cdot3\cdot9^2}=\dfrac{1}{3^7}\)
\(\left(\dfrac{1}{3}\right)^2\cdot\dfrac{1}{3\cdot9^2}=\dfrac{1}{9\cdot3\cdot9^2}=\dfrac{1}{3^7}\)
B1:Tính
a)(-2/3+3/7):4/5+(-1/3+4/7):4/5
b)5/9:(1/11-5/22)+5/9:(1/15-2/3)
c)4^2.4^3/4^10
d)(0,6)^5/(0,2)^6
e)2^7.9^3/6^5.8^2
g)6^3+3.6^2+3^3/-13
h)(3/7+1/2)^2
i)(3/4-5/6)^2
k)5^4.20^44/25^5.4^5
l)(-10/3)^5.(-6/5)^4
f)(1+2/3-1/4).(4/5-3/4)^2
m)2:(1/2-2/3)^3
n)9.9.(-1/3)^3+1/3
y)(4.2^5):(2^3.1/16)
cho biết : 1^2 +2^2+3^2+...+10^2=385
tính A=3^2+6^2+9^2+...+30^2
cho biết : 1^3+2^3+3^3+...+10^3=3025
tính B=2^3+4^3+6^3+...+20^3
(mình sẽ ko tick cho nhg kết quả vd:b đó làm đc câu 1 câu 2 b đó ghi tương tự xin cảm ơn)
cho Sn=\(\frac{1^2-1}{1}+\frac{2^2-1}{2^2}+\frac{3^3-1}{3^3}+...+\frac{n^2-1}{n^2}\)
CMR Sn không phải là số nguyên
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 \(\dfrac{3}{4}\)
Chứng tỏ:
\(\dfrac{1}{3} + \dfrac{1}{3^2} +...+ \dfrac{1}{3^{99}} < \dfrac{1}{2}\)
\(\dfrac{3}{1^2.2^2} + \dfrac{5}{2^2.3^2} + ...+ \dfrac{19}{9^2.10^2}<1 \)
A= 1-|2x-3|
B= |x-3|-|5-x|
C=4-|5x+2|-|3y+12|
D=5-3(x+1)^2
E=1/2(x+1)^2+3
F=15/4+3|x-1|
G=2021/2020+2019(x-2108)^2
Tính nhanh :
1)A=\(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-...+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)
2)B=\(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+...+\frac{1}{2^{99}}\)
3)C=\(\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-\frac{1}{2^{10}}+...-\frac{1}{2^{58}}\)
tính:
a) \(\left(2^{-1}+3^{-1}\right):\left(2^{-1}-3^{-1}\right)+\left(2^{-1}.2^0\right):2^3\)
b) \(\left(-\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
c) \(\left[\left(0,1\right)^2\right]^0+\left[\left(\dfrac{1}{7}\right)^{-1}\right]^2.\dfrac{1}{49}.\left[\left(2^2\right)^3:2^5\right]\)
giúp mk vs các bn mk dg cần gấp lm
Tính : D = 1/3 + 1/3^2 + 1/3^3 + ... + 1/3^2024