CMR
\(32^{17}+16^{21}-2^{82}⋮44\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Ta có:
\(7^{2006}-7^{2005}+7^{2004}\)
\(=7^{2004}\left(7^2-7+1\right)\)
\(=7^{2004}\times43\)
\(\Rightarrow7^{2006}-7^{2005}+7^{2004}\)chia hết cho 43 (vì có chứa thừa số 43)
b) Ta có:
\(32^{17}+16^{21}-2^{82}\)
\(=\left(2^5\right)^{17}+\left(2^4\right)^{21}-2^{82}\)
\(=2^{85}+2^{84}-2^{82}\)
\(=2^{82}\left(2^3+2^2-1\right)=2^{82}\times11=2^{80}\times2^2\times11\)
\(=2^{80}\times44\)
\(\Rightarrow32^{17}+16^{21}-2^{82}\)chia hết cho 44 (vì có chứa thừa số 44)
a, (-17) + 21 + 79 + 17
= (-17 + 17) + (21 + 79)
= 0 + 100
= 100
b, 40 + 22 + (-16) + (-44)
= 40 + {(-16) + (-44} + 22
= 40 - 60 + 22
= - 20 + 22
= 2
c, (-12) + (-47) + (-28) + 47
= - (12 + 28) + (-47 + 47)
= - 40 + 0
= - 40
d, (- 5) + (-3) + 35 + (-17)
= (-5 + 35) - (3 + 17)
= 30 - 20
= 10
b,
= (4,5 x16 - 1,7) : (4,5 x 15 - 1,7)
= 1
c,
= ( 82 - 82 ) : 36 x ( 32 + 17 + 99 - 81 + 1 )
= 0 : 36 x ( 32 + 17 + 99 - 81 + 1 )
= 0 x ( 32 + 17 + 99 - 81 + 1 )
= 0
d,
= (30 : 7,5 + 0,5 x 3 - 1,5) x (4,5 - 4,5)
= (30 : 7,5 + 0,5 x 3 - 1,5) x 0
= 0
1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21+22+23+24+25+26+27+28+29+30+31+32+33+34+35+36+37+38+39+40+41+42+43+44+45+46+47+48+49+50=1275
\(P=\dfrac{32}{64}\cdot\dfrac{-57}{19}+\dfrac{35}{21}\cdot\dfrac{22}{44}=\dfrac{1}{2}\left(-3+\dfrac{5}{3}\right)=\dfrac{1}{2}\cdot\dfrac{-4}{3}=\dfrac{-2}{3}\)
\(Q=\dfrac{75}{125}\cdot\dfrac{82}{164}+\dfrac{49}{98}\cdot\dfrac{-35}{105}=\dfrac{1}{2}\left(\dfrac{3}{5}-\dfrac{1}{3}\right)=\dfrac{1}{2}\cdot\dfrac{4}{15}=\dfrac{2}{15}\)
Lời giải chi tiết:
82 < 86 | 74 < 80 | 17 = 10 + 7 |
95 > 91 | 62 > 59 | 76 > 50 + 20 |
55 < 57 | 44 < 55 | 16 < 12 + 5 |
Ta có : \(32^{17}+16^{21}-2^{82}=\left(2^5\right)^{17}+\left(2^4\right)^{21}-2^{82}=2^{85}+2^{84}-2^{82}\)
\(=2^{80}.2^5+2^{80}.2^4-2^{80}.2^2=2^{80}.\left(2^5+2^4-2^2\right)=2^{80}.44⋮44\)
Vậy \(32^{17}+16^{21}-2^{82}⋮44\left(đpcm\right)\)