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30 và 45
Ta có: \(30=2\cdot3\cdot5;45=3^2\cdot5\)
Do đó: ƯCLN(30;45)\(=3\cdot5=15\)
=>ƯC(30;45)=Ư(15)={1;-1;3;-3;5;-5;15;-15}
42 và 70
TA có: \(42=2\cdot3\cdot7;70=2\cdot7\cdot5\)
Do đó: ƯCLN(42;70)\(=2\cdot7=14\)
=>ƯC(42;70)=Ư(14)={1;-1;2;-2;7;-7;14;-14}
50 và 60
Ta có: \(50=2\cdot5^2;60=2^2\cdot3\cdot5\)
Do đó: ƯCLN(50;60)\(=2\cdot5=10\)
=>ƯC(50;60)=Ư(10)={1;-1;2;-2;5;-5;10;-10}
Giải
Theo đề bài ta có:
abc + ab + a = 874
( 100a + 10b + c ) + ( 10a + b ) + a = 874
111a + 11b + c = 874 ( 1 )
Từ ( 1 ) suy ra 6 < a < 8
Vậy a = 7
Thay a = 7 vào ( 1 ) ta được:
11b + c = 874 – 777 = 97 ( 2 )
Từ ( 2 ) suy ra 7 < b < 9
Vậy b = 8
Thay b = 8 vào ( 2 ) ta được:
88 + c = 97
c = 97 – 88 = 9
Vậy a = 7, b = 8, c = 9
Ta có:
abc + ab + a = 874
789 + 78 + 7 = 874
nhớ bấm giúp mình nha bạn...
ta có :
2017 :2018 = 0,9995044598612488
mả 2018 : 2017 = 1,00049578520526
suy ra 2017 / 2018 < 2018 / 2017
Bạn Nguyễn Quang Kiên trả lời sai rồi , ở kia là số mũ chứ đâu phải phân số đâu , sao làm vậy được
bên góc trái phía dưới câu trả lời có đó bạn:
\(A=1+4+4^2+....+4^{50}\)
\(A=1\left(1+4\right)+4^2\left(1+4\right)+....+4^{49}\left(1+4\right)\)
\(\Rightarrow A=5\left(1+4^2+...+4^{49}\right)\)
\(\Rightarrow A:20\)dư1
Vì 20\(⋮5\)
VÀ chia cho\(1+4^2+....+4^{99}\)
dư 1 \(\Rightarrow A:20dư1\)
Ta có:
\(A=1+4+4^2+...+4^{50}\)
\(\Rightarrow A=1+\left(4+4^2\right)+\left(4^3+4^4\right)+...+\left(4^{49}+4^{50}\right)\)
\(\Rightarrow A=1+20+4^2.\left(4+4^2\right)+...+4^{48}.\left(4+4^2\right)\)
\(\Rightarrow A=1+20+4^2.20+...+4^{48}.20\)
\(\Rightarrow A=1+20.\left(1+4^2+...+4^{48}\right)\)
Vì \(20⋮20\Rightarrow20.\left(1+4^2+...+4^{48}\right)⋮20\)
\(\Rightarrow A:20\)dư 1
Vậy \(A:20\)dư 1
2
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