so sanh = cach hop li
\(\dfrac{4^{100^{ }}+2}{4^{98^{ }}-1}va\dfrac{4^{98^{ }}+2}{4^{96^{ }}-1}\)
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so sanh = cach hop li
a, \(\dfrac{7}{15}va\dfrac{20}{39}\)
b,\(\dfrac{14}{41}va\dfrac{17}{53}\)
c, \(\dfrac{4^{100^{ }}+2}{4^{98}-1}va\dfrac{4^{98^{ }}+2}{4^{96}-1}\)
1) tim so tu nhien x,biet
A)(x-35)-120=0
B)124+(118-x)=217
C)156-(x+61=82
2) tinh nham bang cach them vao o so nay.bớt di cua so hang kia cung mot so tjich hop
Vi du:57+96=(57-4)+(96+4)=53+100=153
Hay tinh nham 35+98
46+29
3)tinh nham bang cay them vao so tru va so bi tru cung mot so thich hop
Vi du:135-98=(135+2)-(98+2)=137-100=37Hay tinh nham 321-96
1354-997
(x-35)-120=0
=>x-35 = 0+120
=>x-35=120
=>x=120+35
=>x=155
tíc mình nha
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2 tháng 7 2021 lúc 21:10
Tính giá trị biểu thức A , biết rằng A M : N Mà M dfrac{dfrac{1}{99}+dfrac{2}{98}+dfrac{3}{97}+dfrac{4}{96}+...+dfrac{97}{3}+dfrac{98}{2}+dfrac{99}{1}}{dfrac{1}{2}+dfrac{1}{3}+dfrac{1}{4}+dfrac{1}{5}+dfrac{1}{6}+...+dfrac{1}{100}} N dfrac{92-dfrac{1}{9}-dfrac{2}{10}-dfrac{3}{11}-...-dfrac{90}{98}-dfrac{91}{99}-dfrac{92}{100}}{dfrac{1}{45}+dfrac{1}{50}+dfrac{1}{55}+...+dfrac{1}{495}+dfrac{1}{500}}
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Tính giá trị biểu thức A , biết rằng A = M : N
Mà M = \(\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
N = \(\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
Ta có: \(M=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(1+\dfrac{2}{98}\right)+\left(1+\dfrac{3}{97}\right)+\left(1+\dfrac{4}{96}\right)+...+\left(1+\dfrac{98}{2}\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{98}+\dfrac{100}{97}+...+\dfrac{100}{1}+\dfrac{100}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
=100
Ta có: \(N=\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
\(=\dfrac{\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{90}{98}\right)+\left(1-\dfrac{91}{99}\right)+\left(1-\dfrac{92}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)
\(=\dfrac{\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{99}+\dfrac{8}{100}}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)
\(=\dfrac{8}{\dfrac{1}{5}}=40\)
\(\Leftrightarrow\dfrac{M}{N}=\dfrac{100}{40}=\dfrac{5}{2}\)
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a. So sanh 2 phan so:A= 2015/2016+2016/2017+2017/2018 va B = 2015+2016+2017/2016+2017+2018
b.1/2.4+1/4.6+........+1/(2x-2).2x = 1/8
c.Cho A = 1/4+1/9+1/16+...+1/81+1/100 . Chung minh rang : A > 65/132
d.Cho B = 12/(2 . 4 ) ^ 2 + 20/ (4 . 6) ^2 + ...........+ 388/ ( 96 . 98 ) ^ 2 + 396/ ( 98 . 100 ) ^2 .Hay so sanh B voi 1 /4
1, so sanh = cach hop li
a,\(\dfrac{31}{67}va\dfrac{29}{73}\)
a) Ta có:\(\dfrac{31}{67}>\dfrac{31}{73}\) (1)
\(\dfrac{31}{73}>\dfrac{29}{73}\) (2)
Từ (1) và (2) \(\Rightarrow\) \(\dfrac{31}{67}>\dfrac{31}{73}>\dfrac{29}{73}\)
\(\Rightarrow\dfrac{31}{67}>\dfrac{29}{73}\)
Vậy:...............
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Rút gọn: A = \(\dfrac{17^{100}+17^{96}+17^{92}+...+17^4+1}{17^{102}+17^{100}+17^{98}+...+17^2+1}\)
\(A=\dfrac{17^{100}+17^{96}+17^{92}+....+17^4+1}{17^{102}+17^{100}+17^{98}+....+17^2+1}\)
Gọi \(17^{100}+17^{96}+17^{92}+....+17^4+1\) là B
\(B=17^{100}+17^{96}+17^{92}+....+17^4+1\\ 17^4\cdot B=17^{104}+17^{100}+17^{96}+......+17^8+17^4\\ 17^4\cdot B-B=\left(17^{104}+17^{100}+17^{96}+......+17^8+17^4\right)-\left(17^{100}+17^{96}+17^{92}+....+17^4+1\right)\\ B\cdot\left(17^4-1\right)=17^{104}-1\\ B=\dfrac{17^{104}-1}{17^4-1}\)
Gọi \(17^{102}+17^{100}+17^{98}+....+17^2+1\) là C
\(C=17^{102}+17^{100}+17^{98}+....+17^2+1\\ C\cdot17^2=17^{104}+17^{102}+17^{100}+17^{98}+....+17^2\\ C\cdot17^2-C=\left(17^{104}+17^{102}+17^{100}+17^{98}+....+17^2\right)-\left(17^{102}+17^{100}+17^{98}+....+17^2+1\right)\\ C\cdot\left(17^2-1\right)=17^{104}-1\\ C=\dfrac{17^{104}-1}{17^2-1}\)
=>
\(A=B:C\\ A=\dfrac{17^{104}-1}{17^4-1}:\dfrac{17^{104}-1}{17^2-1}\\ A=\dfrac{17^2-1}{17^4-1}\)
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A = \(\dfrac{T}{M}\)
M = T + (\(17^{102}+17^{98}+17^{94}+...+17^2\))
M = T + \(17^2\left(17^{100}+17^{96}+17^{92}+...+17^0\right)\)
M = T + \(17^2\cdot\) T = T(\(1+17^2\))
=> A = \(\dfrac{T}{T\left(1+17^2\right)}=\dfrac{1}{1+17^2}\)
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so sanh A va B voi 0
A = 1. ( - 2 ) . 3 . ( - 4 ) ..... . 99 . ( -100 )
B = 1 . ( - 2 ) . 3 . ( - 4 ) . ... . (-98) . 99
A có 50 thừa số âm
=> A > 0
b) CÓ 49 thừa số âm
=> B < 0
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A có 50 thừa số âm
=> A > 0
B có 49 thừa số âm
=> B < 0
tick nha
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so sanh A va B voi 0
A = 1 . ( - 2 ) . 3 .( - 4 ) . .... . 99 . ( - 100 )
B = 1 . ( - 2 ) . 3. ( - 4 ) . .... . ( - 98 ) . 99
Tính : \(\dfrac{x-1}{98}+\dfrac{x-2}{97}+\dfrac{x-3}{96}=\dfrac{x-4}{93}+2\)
mình thấy đề hơi sai sai sao á bạn đoạn \(\dfrac{x-4}{93}\) á bạn
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