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a) \(2010^{100}+2010^{99}\)
\(=2010^{99}\left(2010+1\right)\)
\(=2010^{99}.2011⋮2011\left(dpcm\right)\)
b) \(3^{1994}+3^{1993}-3^{1992}\)
\(=3^{1992}\left(3^2+3-1\right)\)
\(=3^{1992}.11⋮11\left(dpcm\right)\)
c) \(4^{13}+32^5-8^8\)
\(=\left(2^2\right)^{13}+\left(2^5\right)^5-\left(2^3\right)^8\)
\(=2^{26}+2^{25}-2^{24}\)
\(=2^{24}\left(2^2+2-1\right)\)
\(=2^{24}.5⋮5\left(dpcm\right)\)
\(\frac{1}{2}.2^n+4.2^n=9.2^5\Rightarrow2^n\left(\frac{1}{2}+4\right)=288\Rightarrow2^n.\frac{9}{2}=288\Rightarrow2^{n-2}.9=288\Rightarrow2^{n-2}=32\)(dấu "=>" số 3 bn sửa thành 2n-1.9=288=>2n-1=32 nha)
=>2n-1=25=>n-1=5=>n=5+1=6
vậy......
~~~~~~~~~~~~~~~
c, \(\frac{-32}{-2^n}=4\)
\(\Rightarrow-2^n=-32:4\)
\(\Rightarrow-2^n=-8\)
\(\Rightarrow-2^n=-2^3\Rightarrow n=3\)
d, \(\frac{8}{2^n}=2\)
\(\Rightarrow2^n=8:2\)
\(\Rightarrow2^n=4\)
\(\Rightarrow2^n=2^2\Rightarrow n=2\)
e, \(\frac{25^3}{5^n}=25\)
\(\Rightarrow5^n=25^3:25\)
\(\Rightarrow5^n=25^2\)
\(\Rightarrow5^n=5^4\Rightarrow n=4\)
i , \(8^{10}:2^n=4^5\)
\(\Rightarrow2^n=8^{10}:4^5\)
\(\Rightarrow2^n=\left(2^3\right)^{10}:\left(2^2\right)^5\)
\(\Rightarrow2^n=2^{30}:2^{10}\)
\(\Rightarrow2^n=2^{20}\Rightarrow n=20\)
k, \(2^n.81^4=27^{10}\)
\(\Rightarrow2^n=27^{10}:81^4\)
\(\Rightarrow2^n=\left(3^3\right)^{10}:\left(3^4\right)^4\)
\(\Rightarrow2^n=3^{30}:3^{16}\)
\(\Rightarrow2^n=3^{14}\)
\(\Rightarrow2^n=4782969\)Không chia hết cho 2 nên ko có Gt n thỏa mãn
a.
165 + 215 = (24)5 + 215 = 220 + 215 = 215 x (25 + 1) = 215 x (32 + 1) = 215 x 33
Vậy 1615 + 215 chia hết cho 33
b.
817 - 279 - 913 = (34)7 - (33)9 - (32)13 = 328 - 327 - 326 = 322 x (36 - 35 - 34) = 322 x 405
Vậy 817 - 279 - 913 chia hết cho 405
a) \(2010^{100}\)+ \(2010^{99}\)
= \(2010^{99}\)\(\left(2010+1\right)\)
= \(2010^{99}\). \(2011\)chia hết cho 2011
Vậy ...................................
b) \(3^{1994}\)+ \(3^{1993}\)- \(3^{1992}\)
= \(3^{1992}\)\(\left(3^2+3-1\right)\)
= \(3^{1992}\). \(11\)
Vậy .......................
c) \(4^{13}\)+ \(32^5\)- \(8^8\)
= \(\left(2^2\right)^{13}\)+ \(\left(2^5\right)^5\)- \(\left(2^3\right)^8\)
= \(2^{26}\)- \(2^{25}\)- \(2^{24}\)
= \(2^{24}\). \(\left(2^2+2-1\right)\)
= \(2^{24}\). \(5\)
Vậy .......................
3 cau 3 nhe
a)
\(=2010^{99}\left(2010+1\right)\)
\(=2010^{99}.2011\)
cung thay chia het ro nhi
b)
\(=3^{1992}\left(3^2+3-1\right)\)
\(=3^{1992}.11\)
cung thay chia het ro nhi
c)
\(=\left(2^2\right)^{13}+\left(2^5\right)^5-\left(2^3\right)^8\)
\(=2^{26}+2^{25}-2^{24}\)
\(=2^{24}\left(2^2+2-1\right)\)
\(=2^{24}.5\)
cung thay chia het ro nhi
cho 3 nhe
1) 3^1994+4^1993-3^1992
= 3^1992.(9+3-1)=3^1992.11 chia hết cho 11
=> 3^1994+3^1993-3^1992 chia hết cho 11
Có ai bt bài 2 ko z
B2
a)2^n =4.32=128=2^7
=>n=7
b)3^3n.3^2n =9^25
=>3^5n=3^50 =>n=10
bn có thể giải chi tuêts hơn đc ko , mk ko hỉu
1) Chứng minh
a) \(3^{1994}+3^{1993}-3^{1992}⋮11\)
Giải
Ta có : \(3^{1994}+3^{1993}-3^{1992}\)
\(=3^{1992}.\left(3^2+3-1\right)\)
\(=3^{1992}.11⋮11\)
\(\Rightarrow3^{1994}+3^{1993}-3^{1992}⋮11\left(\text{đpcm}\right)\)
b) 413 + 325 - 88 \(⋮\)5
Giải
Ta có : \(4^{13}+32^5-8^8=\left(2^2\right)^{13}+\left(2^5\right)^5-\left(2^3\right)^8\)
\(=2^{2.13}+2^{5.5}-2^{3.8}\)
\(=2^{26}+2^{25}-2^{24}\)
\(=2^{14}.\left(2^2+2-1\right)\)
\(\Rightarrow2^{14}.5⋮5\)
=> 413 + 325 - 88 \(⋮\)5 (ĐPCM)
2) Tìm n biết :
\(a)\frac{2^n}{32}=4\)
\(\Rightarrow\frac{2^n}{2^5}=2^4\)
\(\Rightarrow2^n=2^5.2^4\)
\(\Rightarrow2^n=2^9\)
\(\Rightarrow n=9\)
b) \(27^n.9^n=9^{27}:81\)
\(\Rightarrow\left(27.9\right)^n=9^{27}:9^2\)
\(\Rightarrow\left(3^3.3^2\right)^n=9^{29}\)
\(\Rightarrow3^{5n}=\left(3^2\right)^{29}\)
\(\Rightarrow3^{5n}=3^{58}\)
\(\Rightarrow5n=58\)
\(\Rightarrow n=\frac{58}{5}\)
1.Chứng minh
a) 3^{1994}+3^{1993}-3^{1992}⋮1131994+31993−31992⋮11
giải
Ta có : 3^{1994}+3^{1993}-3^{1992}31994+31993−31992
=3^{1992}.\left(3^2+3-1\right)=31992.(32+3−1)
=3^{1992}.11⋮11=31992.11⋮11
\Rightarrow3^{1994}+3^{1993}-3^{1992}⋮11\left(\text{đpcm}\right)⇒31994+31993−31992⋮11(đpcm)
b) 413 + 325 - 88 ⋮⋮5
Giải
Ta có : 4^{13}+32^5-8^8=\left(2^2\right)^{13}+\left(2^5\right)^5-\left(2^3\right)^8413+325−88=(22)13+(25)5−(23)8
=2^{2.13}+2^{5.5}-2^{3.8}=22.13+25.5−23.8
=2^{26}+2^{25}-2^{24}=226+225−224
=2^{14}.\left(2^2+2-1\right)=214.(22+2−1)
\Rightarrow2^{14}.5⋮5⇒214.5⋮5
=> 413 + 325 - 88 ⋮⋮5 (ĐPCM)
2 Tìm .n biết :
a)\frac{2^n}{32}=4a)322n=4
\Rightarrow\frac{2^n}{2^5}=2^4⇒252n=24
\Rightarrow2^n=2^5.2^4⇒2n=25.24
\Rightarrow2^n=2^9⇒2n=29
\Rightarrow n=9⇒n=9
b) 27^n.9^n=9^{27}:8127n.9n=927:81
\Rightarrow\left(27.9\right)^n=9^{27}:9^2⇒(27.9)n=927:92
\Rightarrow\left(3^3.3^2\right)^n=9^{29}⇒(33.32)n=929
\Rightarrow3^{5n}=\left(3^2\right)^{29}⇒35n=(32)29
\Rightarrow3^{5n}=3^{58}⇒35n=358
\Rightarrow5n=58⇒5n=58
\Rightarrow n=\frac{58}{5}⇒n=558