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Đề ???
\(A=\frac{1003+1007+\frac{2010}{113}+\frac{2010}{117}-\frac{1003}{119}-\frac{1007}{119}}{1003+1008+\frac{2011}{113}+\frac{2011}{117}-\frac{1003}{119}-\frac{1008}{119}}\)
\(=\frac{2010+\frac{2010}{113}+\frac{2010}{117}-\frac{2010}{119}}{2011+\frac{2011}{113}+\frac{2011}{117}-\frac{2011}{119}}\)
\(=\frac{2010.\left(1+\frac{1}{113}+\frac{1}{117}-\frac{1}{119}\right)}{2011.\left(1+\frac{1}{113}+\frac{1}{117}-\frac{1}{119}\right)}\)
\(=\frac{2010}{2011}\)
\(A=\frac{1003+1007+\frac{2010}{113}+\frac{2010}{117}-\frac{100}{119}-\frac{1007}{119}}{1003+1008+\frac{2011}{113}+\frac{2011}{117}-\frac{1003}{119}-\frac{1008}{119}}\)
\(A=\frac{1003+1008+\frac{2011}{113}+\frac{2011}{117}-\frac{1003}{119}-\frac{1008}{119}}{1003+1008+\frac{2011}{113}+\frac{2011}{117}-\frac{1003}{119}-\frac{1008}{119}}\)+ \(\frac{1+\frac{1}{113}+\frac{1}{117}-\frac{903}{119}-\frac{1}{119}}{1003+1008+\frac{2011}{113}+\frac{2011}{117}-\frac{1003}{119}-\frac{1008}{119}}\)
\(A=1+\frac{1+\frac{1}{113}+\frac{1}{117}-\frac{904}{119}}{2011+\frac{2011}{113}+\frac{2011}{117}-\frac{2011}{119}}\)
\(A=\frac{1+\frac{1}{113}+\frac{1}{117}-\frac{1}{119}-\frac{90.}{119}}{2011+2011.\left(\frac{1}{113}+\frac{1}{117}-\frac{1}{119}\right)}\)
\(A=\frac{\frac{90}{119}}{2010+2011}\)
\(A=\frac{\frac{90}{119}}{4021}\)
a) ta có: \(M=1+3+3^2+3^3+...+3^{119}\)
\(M=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{117}+3^{118}+3^{119}\right)\)
\(M=\left(1+3+3^2\right)+3^3.\left(1+3+3^2\right)+...+3^{117}.\left(1+3+3^2\right)\)
\(M=\left(1+3+3^2\right).\left(1+3^3+...+3^{117}\right)\)
\(M=13.\left(1+3^3+...+3^{117}\right)⋮13\left(đpcm\right)\)
b) ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{2010^2}< \frac{1}{2009.2010}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2010^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2009.2010}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(=1-\frac{1}{2010}< 1\)
\(\Rightarrow N=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2010^2}< 1\left(đpcm\right)\)
a, \(M=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{117}+3^{118}+3^{119}\right)\)
\(=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{117}\left(1+3+3^2\right)\)
\(=\left(1+3+3^2\right)\left(1+3^3+3^6+...+3^{117}\right)\)
\(=13.\left(1+3^3+...+3^{117}\right)⋮13\)
b, \(N=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{2010.2010}\)
\(< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2009.2010}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(=1-\frac{1}{2010}=\frac{2009}{2010}< 1\)
\(\Rightarrow N< 1\)
a) 3 . 52 - 16 . 22
= 3 . 25 - 16 . 4
= 75 - 64
= 11
b) (49 . 42 - 47 . 42 ) : 47
= [ 42 . (49 - 47 ) ] : 47
= 42 . 2 : 47
= 2
c) 2448 : [ 119 - ( 23 - 6 ) ]
= 2448 : [ 119 - 17 ]
= 2448 : 102
= 24
d) 25 .37 + 37 . 75 - 270
= 37 . ( 25 + 75 ) - 270
= 37 . 100 - 270
= 3700 - 270
= 3430
e) 5 . 32 - 32 : 22
= 5 . 9 - 32 : 4
= 45 - 8
= 37
g) 18 . [ 270 : ( 15 - 12 )2 ]
= 18 . [ 270 : 32 ]
= 18 . [ 270 : 9 ]
= 18 . 30
= 540
Câu 1 Tính
\(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{20}+...+\frac{1}{2352}+\frac{1}{2450}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{4.5}+...+\frac{1}{48.49}+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{48}-\frac{1}{49}+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}=\frac{49}{50}\)
Câu 2 Tính
\(P=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{99}\right)\left(1-\frac{1}{100}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{98}{99}.\frac{99}{100}\)
\(=\frac{1.2.3...98.99}{2.3.4...99.100}=\frac{1}{100}\)
Câu 3
a) Ta có : M = 1 + 3 + 32 + 33 + ... + 3118 + 3119 (1)
=> 3M = 3 + 32 + 33 + 34 + ... + 3119 + 3120 (2)
Lấy (2) trừ (1) theo vế ta có :
3M - M = (3 + 32 + 33 + 34 + ... + 3119 + 3120) - ( M = 1 + 3 + 32 + 33 + ... + 3118 + 3119)
=> 2M = 3120 - 1
=> M = \(\frac{3^{120}-1}{2}\)
b) M = 1 + 3 + 32 + 33 + ... + 3118 + 3119
= (1 + 3 + 32) + (33 + 34 + 35) + ... + (3117 + 3118 + 3119)
= (1 + 3 + 32) + 33(1 + 3 + 32) + ... + 3117(1 + 3 + 32)
= 13 + 33.13 + ... + 3117.13
= 13(1 + 33 + ... + 3117) \(⋮\)13
=> M \(⋮\)13
M = 1 + 3 + 32 + 33 + ... + 3118 + 3119
= (1 + 3 + 32 + 33) + (34 + 35 + 36 + 37) + ... + (3116 + 3117 + 3118 + 3119)
= (1 + 3 + 32 + 33) + 34(1 + 3 + 32 + 33) + ... + 3116(1 + 3 + 32 + 33)
= 40 + 34.40 + ... + 3116.40
= 40(1 + 34 + ... + 3116)
= 5.8.(1 + 34 + ... + 3116) \(⋮\)5
4) Tính
A = 2100 - 299 - 298 - ... - 22 - 2 - 1
=> 2A = 2101 - 2100 - 299 - 298 - 22 - 2 - 1
Lấy 2A trừ A theo vế ta có :
2A - A = (2101 - 2100 - 299 - 298 - 22 - 2 - 1) - (2100 - 299 - 298 - ... - 22 - 2 - 1)
=> A = 2101 - 2100 - 2100 + 1
=> A = 2101 - (2100 + 2100) + 1
=> A = 2101 - 2100 . 2 + 1
=> A = 1
Câu 5 a) C = 1.2 + 2.3 + 3.4 + ... + 99.100
=> 3C = 1.2.3 + 2.3.3 + 3.4.3 + .... + 99.100.3
= 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98)
= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100
= 99.100.101
=> C = 99.100.101 : 3 = 333300
b) Ta có : D = 22 + 42 + 62 + ... + 982
= 22(12 + 22 + 32 + ... + 492
= 22 .(12 + 22 + 32 + ... + 492)
= 22.(1.1 + 2.2 + 3.3 + ... + 49.49)
= 22.[1.(2 - 1) + 2..(3 - 1) + 3(4 - 1) + ... + 49(50 - 1)]
= 22.[(1.2 + 2.3 + 3.4 + ... + 49.50) - (1 + 2 + 3 + 4 + ... + 49)]
Đặt E = 1.2 + 2.3 + 3.4 + ... + 49.50
=> 3E = 1.2.3 + 2.3.3 + 3.4.3 + .... + 49.50.3
= 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 49.50.(51 - 48)
= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 49.50.51 - 48.49.50
= 49.50.51
=> E = 49.50.51/3 = 41650
Khi đó D = 22.[41650 - (1 + 2 + 3 + 4 + ... + 49)]
= 22.[41650 - 49(49 + 1)/2]
= 22.[41650 - 1225
= 22.40425
= 161700
=> D = 161700
M=0:(119.119 + 2024) + (119:119 - 0.2024)
= 0 + (1 - 0 )
= 1
Tính giá trị của biểu thức M:
Với a=119 và b=0, ta có:
M=b:(119⋅a+2024)+(119:a−b⋅2024)
Thay a=119 và b=0 vào biểu thức:
M=0:(119⋅119+2024)+(119:119−0⋅2024)
M=0+(1−0)
M=1
Kết quả: M=1