Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(\dfrac{1}{2}\) - ( - \(\dfrac{1}{3}\) ) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= \(\dfrac{5}{6}\) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= 1 + \(\dfrac{1}{23}\)
= \(\dfrac{24}{23}\)
b, \(\dfrac{11}{24}\) - \(\dfrac{5}{41}\) + \(\dfrac{13}{24}\) + 0,5 - \(\dfrac{36}{41}\)
= (\(\dfrac{11}{24}\) + \(\dfrac{13}{24}\)) - ( \(\dfrac{5}{41}\) + \(\dfrac{36}{41}\)) + 0,5
= 1 - 1 + 0,5
= 0,5
c,\(-\dfrac{1}{12}-\left(\dfrac{1}{6}-\dfrac{1}{4}\right)\)
=\(-\dfrac{1}{12}-\left(-\dfrac{1}{12}\right)\)
=0
d, \(\dfrac{1}{6}-\left[\dfrac{1}{6}-\left(\dfrac{1}{4}+\dfrac{9}{12}\right)\right]\)
= \(\dfrac{1}{6}-\left[\dfrac{1}{6}-1\right]\)
= \(\dfrac{1}{6}-\left(-\dfrac{5}{6}\right)\)
= 1
A = { 1,2,3,...2017 }
= ( 2017 - 1 ) : 1 + 1
= 2017 ( phần tử )
B = { 24,26 , 28 , ..., 2018 }
= ( 2018 - 24 ) : 2 + 1
= 998 ( phần tử)
^^ Học tốt nhé!
C= { 15,17,19,...,2017 }
= ( 2017 - 15 ) : 2 + 1
= 1002 ( phần tử)
C = \(\frac{25^{28}+25^{24}+...+25^4+25^0}{25^{30}+25^{28}+...+25^2+25^0}\)
= \(\frac{25^{28}+25^{24}+...+25^2+25^0}{\left(25^{30}+25^{26}+...+25^2\right)+\left(25^{28}+25^{24}+...+25^0\right)}\)
= \(\frac{25^{28}+25^{24}+...+25^0}{25^2\left(25^{28}+25^{24}+...+1\right)+\left(25^{28}+25^{24}+...+1\right)}\)
= \(\frac{25^{28}+25^{24}+...+1}{\left(25^2+1\right)\left(25^{28}+25^{24}+...+1\right)}=\frac{1}{626}\)
Cho 10 ****
22.23-(56:54-20100.24)3-14
= 25-(52-1.24)3-14
=32 - (25-24)3-14
=32 - 13-14
=32-1-14=32-15=17
ta có: 1/1*6+1/6*11+1/6*16+...+1/51*56.
=1/5.(5/1.6+5/6.11+5/6.16+...+5/51.56)
=1/5.(1/1-1/6+1/6-...-1/56)
=1/5.(1-1/56)
=1/5.(55/56)
=11/56
-7-2x=37-(-24)
-7-2x=61
-2x=61+7
-2x=68
x=68:(-2)
x=-34
k cho mình nhé
\(B=2+2^2+2^3+2^4+...+2^{99}+2^{100}=2\left(1+2^2+2^3+2^4\right)+...+2^{96}\left(1+2^2+2^3+2^4\right)=2.31+2^6.31+...+2^{96}.31=31\left(2+2^6+...+2^{96}\right)⋮31\)
có phải câu hỏi như thế này ko
155 x (-24)+24x55
ĐÚNG RỒI