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\(Ư\left(144\right)=\left\{1,2,3,4,6,8,9,12,16,18,24,36,48,72,144\right\}\)
\(Ư\left(324\right)=\left\{1,2,3,4,6,9,12,18,27,36,54,81,108,162,324\right\}\)
\(Ư\left(576\right)=\left\{1,2,3,4,6,8,9,12,16,18,24,32,36,48,64,72,96,144,192,288,576\right\}\)
\(Ư\left(1024\right)=\left\{1,2,4,8,16,32,64,128,256,512,1024\right\}\)
\(Ư\left(1296\right)=\left\{1,2,3,4,6,8,9,12,16,24,27,36,48,54,72,81,108,144,162,216,324,432,648,1296\right\}\)
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Câu hỏi:
Ư(81) = {........................}
Trả lời:
Ư(81) = { -1 ; -3 ; -9 ; -27 ; -81 ; 1 ; 3 ; 9 ; 27 ; 81 }.
Học tốt
A = { 0 ; 1 ; 2 ; 3 ; 4 ; .... ; 50 }
B = { 1 ; 3 ; 5 ; 9 ; 15 ; 45 }
C = { 0 ; 15 ; 30 ; 45 ; 60 ; ... }
A \(\Omega\)B = { 1 ; 3 ; 5 ; 9 ; 15 ; 45 }
A \(\Omega\)C = { 0 ; 15 ; 30 ; 45 }
B \(\Omega\)C = { 15 ; 45 }
1)
\(Ư\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
\(Ư\left(14\right)=\left\{\pm1;\pm2;\pm7;\pm14\right\}\)
\(Ư\left(17\right)=\left\{\pm1;\pm17\right\}\)
\(Ư\left(1\right)=\left\{\pm1\right\}\)
2)
a)
\(Ư\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
b)
\(Ư\left(18\right)=\left\{\pm1;\pm2;\pm3;\pm6;\pm9;\pm18\right\}\)
c)
\(Ư\left(24\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm8;\pm12;\pm24\right\}\)
\(\text{Ta có:}\)\(x>8\)\(\Rightarrow\)\(x\in\left\{12;24\right\}\)
Ư(111)\(\in\)\(^{\left\{1;3;37;111\right\}}\)
n+2=1\(\Rightarrow\)n=1-2=-1 \(\notin\)N loại+
n+2=3\(\Rightarrow\)n=3-2=1
n+2=37\(\Rightarrow\)n=37-2=35
n+2=111\(\Rightarrow\)n=111-2=109
Vậy n\(\in\)\(\left\{1;35;109\right\}\)
ta có n+2\(\in\)Ư(111)= (\(\pm\)1, \(\pm\)3, \(\pm\)111)
ta cóa bảng
| n+2 | 1 | -1 | 3 | -3 | 111 | -111 |
| n | -1 | -3 | 1 | -5 | 109 |
-113 |
vậy n......
a) Ư(12²) = Ư(144) = {1; 2; 3; 4; 6; 8; 9; 12; 16; 18; 24; 32; 48; 72; 144}
b) Ư(18²) = Ư(324) = {1; 2; 3; 4; 6; 9; 12; 18; 27; 36; 54; 81; 108; 162; 324}
c) Ư(24²) = Ư(576) = {1; 2; 3; 4; 6; 8; 9; 12; 16; 18; 24; 32; 36; 48; 64; 72; 96; 144; 192; 288; 576}
d) Ư(32²) = Ư(1024) = {1; 2; 4; 8; 16; 32; 64; 128; 256; 512; 1024}