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1.Ta có A= 710 +79 - 78
A= 78 .(72 +7 -1)
A=78 .55
=> A chia hết cho 11( vì có thừa số 55 chia hết cho 11)
Ta thấy: 7 + 72 + 73 + 74 = 7 + 49 + 343 + 2401 = 2800 chia hết cho 202
P = 7 + 72 + 73 + ... + 72016 = ( 7 + 72 + 73 + 74) + 74( 7 + 72 + 73 + 74) + ... + 72012( 7 + 72 + 73 + 74)
P = 2800 + 74 . 2800 + ... + 72012 . 2800 = 2800( 1 + 74 + ... + 72012 )
Mà 2800 chia hết cho 202 \(⇒\) P chia hết cho 202
\(a.\)
\(8^7-2^{18}\)
\(=\left(2^3\right)^7-2^{18}\)
\(=2^{21}-2^{18}\)
\(=2^{18}.2^3-2^{18}\)
\(=2^{18}\left(2^3-1\right)\)
\(=2^{18}.7\)
\(=2^{17}.7.2⋮14\)
Vậy \(8^7-2^{18}⋮14\)
\(b.\)
\(5^5-5^4+5^3\)
\(=5^3\left(5^2-5+1\right)\)
\(=5^3.21\)
\(=5^3.7.3⋮7\)
Vậy \(5^5-5^4+5^3⋮7\)
\(c.\)
\(7^6+7^5-7^4\)
\(=7^4\left(7^2+7-1\right)\)
\(=7^4.55\)
\(=7^4.5.11⋮11\)
Vậy \(7^6+7^5-7^4⋮11\)
a ) 76 + 75 - 74
= 74 ( 72 + 7 - 1 )
= 74. 55 chia hết cho 55
b ) 165 + 215
= ( 24 ) 5 + 215
= 220 + 215
= 215 ( 25 + 1 )
= 215 . 33 chia hết cho 33
c ) 817 - 279 - 913
= ( 34 )7 - ( 33 )9 - ( 32 )13
= 328 - 327 - 326
= 326 ( 32 - 3 - 1 )
= 326 . 5
= 322 . 34 . 5
= 322 . 81 . 5
= 322 . 405 chia hết cho 405
a/ Ta có
97 - 312 = 97 - (32) 6 = 97 - 96 = 96 ( 9 - 1 ) = 96. 8 chia hết cho 8
b/ Ta có
76 + 75 - 74 = 75( 7 + 1 ) - 74 = 75.8 - 74 = 74 . 7 . 8 - 74 = 74 . 56 - 74 = 74 . ( 56 -1 ) = 74 . 55 = 74 . 11 . 5 chia hết cho 11
\(2010^{100}+2010^{99}=2010^{99}.\left(2010+1\right)=2010^{99}.2011\)chia hết cho 2011
a, 2010100+201099=201099(2010+1)=201099.2011 =>2010100+201099 chia hết cho 11
\(\frac{x+6}{15}=\frac{5-x}{7}\)
\(\Leftrightarrow\left(x+6\right).7=\left(5-x\right).15\)
\(\Leftrightarrow7x+42=75-15x\)
\(\Leftrightarrow7x+15x=75-42\)
\(\Leftrightarrow22x=33\)
\(\Leftrightarrow x=\frac{3}{2}\)
=> 7.(x+6)= 15.(5-x)
=> 7x +7.6=15.5-15x
=> 7x + 42= 75 -15x
=> 7x+15x=75-42
=> 22x=33
=>x= 1,5
\(125^7-25^{10}+5^{19}\)
\(=\left(5^3\right)^7-\left(5^2\right)^{10}+5^{19}\)
\(=5^{21}-5^{20}+5^{19}\)
\(=5^{19}.\left(5^2-5+1\right)\)
\(=5^{19}.21\)
\(=5^{18}.5.21\)
\(=5^{18}.105\)
Ta có: \(105⋮105\)
\(\Rightarrow5^{18}.105⋮105\)
\(\Rightarrow125^7-25^{10}+5^{19}⋮105\)
đpcm
\(125^7-25^{10}+5^{19}\)
\(=\left(5^3\right)^7-\left(5^2\right)^{10}+5^{19}\)
\(=5^{21}-5^{20}+5^{19}\)
\(=5^{19}.\left(5^2-5+1\right)\)
\(=5^{19}.21\)
\(=5^{18}.5.21=5^{18}.105⋮105\)
Vậy ......