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Bài 4:
x O y z m n
Giải:
Vì Om là tia phân giác của góc xOz nên:
mOz = 1/2.xOz
Vì On là tia phân giác của góc zOy nên:
zOn = 1/2 . zOy
Ta có: xOz + zOy = 180o ( kề bù )
=> 1/2(xOz + zOy) = 1/2 . 180o
=> 1/2.xOz + 1/2.zOy = 90o
=> mOz + zOn = 90o
=> mOn = 90o (đpcm)
Bài 2:
7^6 + 7^5 - 7^4 = 7^4.( 7^2 + 7 - 1 ) = 7^4 . 55 chia hết cho 55
Vậy 7^6 + 7^5 - 7^4 chia hết cho 55
A = 1 + 5 + 5^2 + ... + 5^50
=> 5A = 5 + 5^2 + 5^3 + ... + 5^51
=> 5A - A = ( 5 + 5^2 + 5^3 + ... + 5^51 ) - ( 1 + 5 + 5^2 + ... + 5^50 )
=> 4A = 5^51 - 1
=> A = ( 5^51 - 1 )/4
Bài 2:
a: \(=7^4\left(7^2+7-1\right)=7^4\cdot55⋮55\)
b: \(5A=5+5^2+...+5^{51}\)
\(\Leftrightarrow4A=5^{51}-1\)
hay \(A=\dfrac{5^{51}-1}{4}\)
Bài 3:
\(S=\left(1^2+2^3+3^3+...+10^2\right)\cdot2=385\cdot2=770\)
b) 817 - 279 -913 chia hết cho 405
Ta có: 817 - 279 -913 = 328- 327-326
= 326(32-3-1)
= 326. 5 = 322. 405 chia hết cho 405 (đpcm)
a, \(7^6+7^5-7^4=7^4\left[7^2+4-1\right]=7^4\cdot55⋮55\)
b, \(A=1+5+5^2+5^3+...+5^{50}\)
\(\Rightarrow5A=5+5^2+5^3+5^4+...+5^{51}\)
\(\Rightarrow5A-A=\left[5+5^2+5^3+5^4+...+5^{51}\right]-\left[1+5+5^2+5^3+...+5^{50}\right]\)
\(\Rightarrow4A=5^{51}-1\Leftrightarrow A=\frac{5^{51}-1}{4}\)
Giải:
a) Ta có:
\(7^6+7^5-7^4\)
\(=7^4\left(7^2+7-1\right)\)
\(=7^4.55⋮55\)
Vậy ...
b) Ta có:
\(16^5+2^{15}\)
\(=\left(2^4\right)^5+2^{15}\)
\(=2^{20}+2^{15}\)
\(=2^{15}\left(2^5+1\right)\)
\(=2^{15}.33⋮33\)
Vậy ...
c) \(81^7-27^9-9^{13}\)
\(=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\)
\(=3^{28}-3^{27}-3^{26}\)
\(=3^{26}\left(3^2-3-1\right)\)
\(=3^{26}.5⋮5⋮405\)
Vậy ...
Chúc bạn học tốt!
a) 76 +75 -74
=74.72 +74.7-74
=74.(72+7-1)
=74.55⋮55
b) 165+215
=(24)5 +215
=220+215
=215.25+215
=215.(25+1)
=215.33⋮33
c)817-279-913
=(34)7-(33)9......(làm tương tự)
a) ta có : \(7^6+7^5-7^4=7^4\left(7^2+7-1\right)=7^4.\left(49+7-1\right)=7^4.55⋮55\)
\(\Rightarrow7^4.55\) chia hết cho \(55\) \(\Leftrightarrow7^6+7^5-7^4\) chia hết cho \(55\)
vậy \(7^6+7^5-7^4\) chia hết cho \(55\) (đpcm)
b) ta có \(16^5+2^{15}=\left(2^4\right)^5+2^{15}=2^{20}+2^{15}=2^{15}\left(2^5+1\right)=2^{15}.\left(32+1\right)=2^{15}.33⋮33\)
\(\Rightarrow2^{15}.33\) chia hết cho \(33\) \(\Leftrightarrow16^5+2^{15}\) chia hết cho \(33\)
vậy \(16^5+2^{15}\) chia hết cho \(33\) (đpcm)
c) ta có \(81^7-27^9-9^{13}=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}=3^{28}-3^{27}-3^{26}\)
\(=3^{22}\left(3^6-3^5-3^4\right)=3^{22}\left(729-243-81\right)=3^{22}.405⋮405\)
\(\Rightarrow3^{22}.405\) chia hết cho \(405\) \(\Leftrightarrow81^7-27^9-9^{13}\) chia hết cho \(405\)
vậy \(81^7-27^9-9^{13}\) chia hết cho \(405\) (đpcm)
\(a.\)
\(7^6+7^5-7^4=7^4\left(7^2+7-1\right)=7^4.55⋮55\)
\(b.\)
\(16^5+2^{15}=2^{20}+2^{15}=2^{15}\left(2^5+1\right)=2^{15}.33⋮33\)
\(c.\)
Ta có : \(405=3^4.5\)
\(\Rightarrow81^7-27^9-9^{13}=3^{28}-3^{27}-3^{26}=3^{26}\left(3^2-3-1\right)=3^{26}.5⋮405\)
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:\(7^6+7^5-7^4=7^4\left(7^2+7-1\right)=7^4.55\)
=>\(7^6+7^5-7^4⋮55\)
b)\(A=1+5+5^2+...+5^{50}\)
\(5A=5\left(1+5+5^2+...+5^{50}\right)=5+5^2+5^3+...+5^{51}\)
\(5A-A=5+5^2+5^3+...+5^{51}-\left(1+5+5^2+...+5^{50}\right)\)
\(4A=5^{51}-1\)
\(\Rightarrow A=\dfrac{5^{51}-1}{4}\)
a) \(7^6+7^5+7^4=7^4\left(7^2+7+1\right)\)
= \(7^4.55\)
Vậy: \(7^6+7^5+7^4\) chia hết cho 55.
b) A= \(1+5+5^2+5^3+5^4+.....+5^{50}\)
5A= 5+\(5^2+5^3+5^4+5^{51}\)
5A-A= 5+\(5^2+5^3+5^4+......+5^{51}\)\(-\left(1+5^2+5^3+5^4+......+5^{51}\right)\)
4A= 5+\(5^2+5^3+5^4+......+5^{51}\)\(-1-5-5^2-5^3-5^4-.......-5^{50}\)
= \(5^{51}-1\)
Vậy A= \(\left(5^{51}-1\right):4\)
Tick mk nha!