Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Gọi biểu thức trên là A.
A = 3 + 32+33+ ... + 320
3A = 32+33+34+ ... + 321
3A - A = ( 32+33+34+ ... + 321 ) - ( 3 + 32+33+ ... + 320)
3A - A = 321 - 3
Bn làm tp ik nha
b ) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
= 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
c ) Đặt A = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)
=> A < \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
=> A < 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100= 1 - 1/100 = 99/100 < 1
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)< 1
b, \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\)\(\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
c,Ta thấy
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(.....\)
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\left(đpcm\right)\)
\(3^3\cdot5^3-20\cdot\left\{300-\left[546-2^3\cdot\left(7^8-7^6+7^0\right)\right]\right\}\)
\(=3^3\cdot5^3-20\cdot\left\{300-\left[546-2^3\cdot5647153\right]\right\}\)
\(=3^3\cdot5^3-20\cdot\left\{300-\left[546-45177224\right]\right\}\)
\(=3^3\cdot5^3-20\cdot\left\{300--45176678\right\}\)
\(=3^3\cdot5^3-20\cdot45176978\)
\(=3375-903539560\)
\(=-903536185\)
\(626500:\left\{50^2:\left[178-4\cdot\left(35-21:3\right)\right]\right\}\)
\(=626500:\left\{50^2:\left[178-4\cdot\left(35-7\right)\right]\right\}\)
\(=626500:\left\{50^2:\left[178-4\cdot28\right]\right\}\)
\(=626500:\left\{50^2:\left[178-112\right]\right\}\)
\(=626500:\left\{50^2:66\right\}\)
\(=626500:\frac{1250}{33}\)
\(=\frac{82698}{5}\)
CHUC BAN HOC TOT >.<
23.5 -321:319+80
=8.5-32+1
=40-9+1
=31+1=32
còn bài tìm x thì ko có giá trị nào bạn ạ (vì ko có lũy thừa 5 nào tận cùng khác 5 cả)
Trả lời :
a) 43 . 101 - 64
= 64 . 101 - 64
= 64 . ( 101 - 1 )
= 64 . 100 = 6400
b) \(\frac{5^8}{5^5}+3^2.3^3-\left(30-19\right)^2\)
= 53 + 35 - 112
= 125 + 243 - 121
= 247
a) 43 .101 - 64 =
= 43 . 101 - 43
= 43 .( 101 - 1 )
= 43 .100
= 6400
36 x 12 + 36 x 45 + 36 x 33
= 36 x (12 + 45 + 33)
= 36 x 90
= 3240
b) 78 x 31 + 78 x 24 + 78 x 17 + 22 x 72
=78 x (31 + 24 + 17) + 22 x 72
= 78 x 72 + 22 x 72
= (78 + 22) x 72
= 100 x 72
= 7200
c) ( 34 - 17 )2 - 23 x 32
\(=17^2-8.9\)
\(=289-72\)
\(=217\)
a) \(4.5^2-81:3^2=4.25-81:9=100-9=91\)
b) \(3^3.23-3^3.19=3^3.\left(23-19\right)=27.4=108\)
c) \(2^4.5-[131-\left(13-4\right)^2]=16.5-\left(131-9^2\right)=80-\left(131-81\right)=80-50=30\)
d) \(100:\left\{250:\left[450-\left(4.5^3-2^2.25\right)\right]\right\}\)
\(=100:\left\{250:\left[450-4.\left(125-25\right)\right]\right\}\)
\(=100:\left[250:\left(450-4.100\right)\right]\)
\(=100:\left[250:\left(450-400\right)\right]=100:\left(250:50\right)=100:5=20\)
\(a,5:5^2+2^3.3^2\)
\(=5^4+72\)
\(=697\)
\(b,8^3.25-8^3.23\)
\(=8^3.\left(25-23\right)\)
\(=8^3.2\)
\(=1024\)
\(c,80-\left(4.5^2-3^2.2^3\right)\)
\(=80-\left(100-72\right)\)
\(=80-28\)
\(=52\)
\(d,\left(12^4-6^4+2.3^4\right):3^4\)
\(=\left(20736-1296+162\right):3^4\)
\(=19602:3^4\)
\(=242\)
3S = 32 - 33 + 34 - 35 + ... + 320 - 321
3S + S = (32 - 33 + 34 - 35 + ... + 320 - 321) + (3 - 32 + 33 - 34 + ... + 319 - 320)
4S = -321 + 3
S = -321 + 3/4
\(S=3-3^2+3^3-3^4+......+3^{19}-3^{20}\)
\(\Rightarrow3S=3^2-3^3+3^4-.........-3^{19}+3^{20}-3^{21}\)
\(\Rightarrow3S+S=\left(3^2-3^3+3^4-.......-3^{19}+3^{20}-3^{21}\right)+\left(3-3^2+3^3-....+3^{19}-3^{20}\right)\)
\(\Rightarrow4S=-3^{21}+3\)
\(\Rightarrow S=\frac{-3^{21}+3}{4}\)