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Gọi hai số tự nhiên cần tìm là a và b (a ; b thuộc N )
Vì ƯCLN ( a, b ) = 36 nên a = 36 m ; b = 36n
ƯCLN(m , n ) = 1
Theo đề bài ra , ta có : a + b = 36m + 36n = 432 => 36(m+n) = 432 => m + n = 12
=> Ta tìm được các cặp mn thoả mãn điều kiện :
(m,n) = {( 1,11);(11,1);(5,7);(7,5)}
=> (a,b) = {(36, 396);(396;36);(180, 252);(252,180)}
a) goi hai so la a ; b va a >b
vi UCLN(a,b)=18=>a=18k ; b=18q (trong do UCLN (k,q)=1 va k>q)
=>a+b=162
18k+18q =162
18(k+q)=162
k+q=9
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Ta có: \(99^{20}=\left(99^2\right)^{10}=9801^{10}\)
Vì \(9801< 9999\)nên \(9801^{10}=9999^{10}\)
Vậy \(99^{20}< 9999^{10}\)
a, 2^27 = 2^3.9 = 8^9
3^18 = 3^2.9 = 9^9
vì 8<9 => 8^9 < 9^9 => 2^27 < 3^18
\(2^{50}=\left(2^5\right)^{10}=32^{10}\)
\(5^{20}=\left(5^2\right)^{10}=25^{10}\)
Suy ra: 250 > 520
b)
\(9^{200}=\left(9^2\right)^{100}=81^{100}\)
Suy ra: 99100 > 81100
a) 36 = 2 . 2 . 3 . 3
52 = 2 . 2 . 13
=> UCLN ( 36 ; 52 ) = 2
=> UC ( 36 ; 52 ) = { \(\pm\)1 ; \(\pm\)2 }
+) 18 = 2 . 3 . 3
63 = 3 . 3 . 7
99 = 3 . 3 . 11
=> UCLN ( 18 ; 63 ; 99 ) = 3
=> UC ( 18 ; 63 ; 99 ) = { \(\pm\)1 ; \(\pm\)3 }
b) 35 = 5 . 7
125 = 5 .5 .5
=> UCLN ( 35 ; 125 ) = 5
=> UC ( 35 ; 125 ) = { \(\pm\)1 ; \(\pm\)5 }
+) 40 = 2 . 2 . 2 . 5
50 = 2 . 5 . 5
100 = 2 . 2 . 5 . 5
=> UCLN ( 40 ; 50 ; 100 ) = 2 . 5 = 10
=> UC ( 40 ; 50 ; 100 ) = { \(\pm\)1 ; \(\pm\)2 ; \(\pm\)5 ; \(\pm\)10 }