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\(A=\frac{4157-19}{12471-57}\)\(=\frac{4138}{12414}\)\(=\frac{4138:4138}{12414:4138}\)\(=\frac{1}{3}\)
\(B=\frac{7}{10^2+8.10^2}\)\(=\frac{7}{100+8.100}\)\(=\frac{7}{100+800}\)\(=\frac{7}{900}\)
\(C=\frac{31995}{42660-108}\)\(=\frac{31995}{42552}\)\(=\frac{31995:27}{42552:27}\)\(=\frac{1185}{1576}\)
\(D=\frac{2^{45}.5^3.2^6.3}{8.2^{18}.81.5}=\frac{2^{51}.5^3.3}{2^3.2^{18}.3^4.5}=\frac{2^{51}.5^3.3}{2^{21}.3^4.5}=\frac{2^{30}.5^2}{3^3}\)
k mình nhé.
A=4138/12414=1/3
B=7/900
C=31995/42552=1185/1576
Phần D tui chịu, ahihi
Ta có : \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)\(=1+\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{50.50}\)
Vì \(\frac{1}{2.2}< \frac{1}{1.2};\frac{1}{3.3}< \frac{1}{2.3};..;\frac{1}{50.50}< \frac{1}{49.50}\)nên :
\(\Rightarrow\) \(1+\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{50.50}\)\(< 1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)
Ta có : \(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(=1+\left(1-\frac{1}{50}\right)\)\(=1+\frac{49}{50}\)
Vì \(\frac{49}{50}< 1\)nên \(1+\frac{49}{50}< 2\)\(\Rightarrow\)\(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}< 2\)
\(\Rightarrow\)\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)\(< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}< 2\)
a) 111 111 + 222 222 + 99 999 + 88 888
= ( 111 111 + 99 999) + ( 222 222 + 88 888)
= 211 110 + 211 110
= 422 220
b) (2x+3).(5y+6) = 55 = 11.5 = (-11).(-5) = 1.55 = (-1).(-55)
TH1: * 2x + 3 = 11 => 2x = 8 => x = 4
5y + 6 = 5 => 5y = -1 => y= -1/5
* 2x + 3 = 5 => 2x = 2 => x= 1
5y + 6 = 11 => 5y = 5 => y = 1
TH2: * 2x+3 = = -11 => 2x = -14 => x= -7
5y + 6 = -5 => 5y = -11 => y= -11/5
* 2x + 3 = -5 => 2x = -8 => x = - 4
5y + 6 = - 11 => 5y = - 17 => y= -17/5
TH3: * 2x +3 = 1 => 2x = -2 => x= -1
5y + 6 = 55 => 5y = -49 => y= -49/5
* 2x + 3 = 55 => 2x = 52 => x = 26
5y + 6 = 1 => 5y = - 5 => y = - 1
TH4: * 2x + 3 = - 1 => 2x = -4 => x= -2
5y + 6 = -55 => 5y = - 61 => y = -61/5
* 2x + 3 = - 55 => 2x = - 58 => x = - 29
5y + 6 = -1 => 5y = -7 => y = -7/5
KL: (x;y) = ............
c) \(\frac{99999^3.88888^2}{99999^2.88888^3}=\frac{99999}{88888}=\frac{11111.9}{11111.8}=\frac{9}{8}\)
d) Gọi Ư C L N ( 2n + 1; 22n +3) là d
ta có: 2n +1 chia hết cho d => 11.( 2n +1) chia hết cho d => 22n + 11 chia hết cho d
22n + 3 chia hết cho d
=> 22n + 11 - 22n -3 chia hết cho d => 7 chia hết cho d
xem lại đề bài!
a) 111111 + 222222 + 99999 + 88888
= 111111 + 99999 + 222222 + 88888
= (111111 + 100000 - 1) + ( 222222 + 100000 - 11112 )
= ( 111111 -1 + 100000 ) + ( 222222 -11112 + 100000)
= 111110 + 100000 + 211110 + 100000
= 211110 +
= 311110
= 522220
c) \(\frac{99999^3.88888^2}{99999^2.88888^3}\)
\(=\frac{99999.1}{1.88888}\)
\(=\frac{9}{8}\)
có vô số cặp
làm 1 vài phép biến đổi có thể suy ra 15a+10b=6a+6b
<=> 11a+4b=0 <=> a=\(\frac{-4b}{11}\) => -4b thuộc bội của (11)={0;±11;±22;±33,....}
hay b thuộc bội của (44)={0;±44;±88;±132;...}
Mỗi giá trị của b lại có 1 giá trị cua a mà B(44) có vô số số hạng nên có vô số cặp số (a;b) tự nhiên.
ta có :
\(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3}\)
\(............................\)
\(\frac{1}{20^2}=\frac{1}{20.20}< \frac{1}{19.20}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+......+\frac{1}{20^2}< \frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{19.20}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..........+\frac{1}{19}-\frac{1}{20}\)
\(\Rightarrow A< 1-\frac{1}{20}\)
\(\Rightarrow A< \frac{19}{20}\) ( 1 )
mà \(\frac{19}{20}< 1\) ( 2 )
từ ( 1 ) và ( 2 ) \(\Rightarrow A< 1\)
\(\Rightarrow\text{Đ}PCM\)
Ta có :
\(A=\frac{2^{18}-3}{2^{20}-3}\) \(B=\frac{2^{20}-3}{2^{22}-3}\)
\(2^2A=\frac{2^{20}-12}{2^{20}-3}\) \(2^2B=\frac{2^{22}-12}{2^{22}-3}\)
\(4A=\frac{2^{20}-3-9}{2^{20}-3}\) \(4B=\frac{2^{22}-3-9}{2^{22}-3}\)
\(4A=1-\frac{9}{2^{20}-3}\) \(4B=1-\frac{9}{2^{22}-3}\)
Ta thấy 220 - 3 < 222 - 3 nên \(\frac{9}{2^{20}-3}>\frac{9}{2^{22}-3}\)hay \(1-\frac{9}{2^{20}-3}>1-\frac{9}{2^{22}-3}\)
=> 4A > 4B hay A > B
Vậy A > B
Ủng hộ mk nha !!! ^_^
\(A=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}....\frac{100^2-1}{100^2}\)
\(A=\frac{1.3}{2^2}.\frac{2.4}{3^2}....\frac{99.101}{100^2}\)
\(A=\frac{1.3.2.4...99.100}{2.2.3.3...100.100}\)
\(A=\frac{1.2...99}{2.3....100}.\frac{3.4...101}{2.3...100}\)
\(A=\frac{1}{100}.\frac{101}{2}\)
\(A=\frac{101}{200}\)
Tính nhanh :
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{18.19}+\frac{2}{19.20}\)
\(=2.\left(\frac{1}{1.2}+\frac{2}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{18.19}+\frac{1}{19.20}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\right)\)
\(=2.\left(1-\frac{1}{20}\right)\)
\(=2.\left(\frac{20}{20}-\frac{1}{20}\right)\)
\(=2.\frac{19}{20}\)
\(=\frac{19}{10}\)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{18.19}+\frac{2}{19.20}\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(=2.\left(1-\frac{1}{20}\right)\)
\(=2.\frac{19}{20}\)
\(=\frac{19}{10}\)
ủng hộ đi
\(\frac{3}{2}\): 99 999 = \(\frac{3}{2}\)X \(\frac{1}{99999}\)= \(\frac{3}{199998}\)