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1. 2x=16\(\Rightarrow\)X=4
2. 22x-1=27
\(\Rightarrow\)27=22.4-1
Vậy x =4
a,( 393+390) : (317. 373)
= (33+1). 390 : 390
= 33+1
=27+1
=28
b,(556+57) : (549+1)
=57. (549+1) : (549+1)
=57= 78125
c,(722+721+720) ; (25+24+32)
= 720. (72+71+1) : [24. (2+1)+32 ]
= 720. 57 : [ 24. 3 +32 ]
= 720. 57 : ( 24+3) . 3
= 720. 57 : 19 . 3
= 720. 57 : 57
= 720
*Ta có: A\(=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(=\left(2+2^2\right)+2^2\times\left(2+2^2\right)+...+2^{2008}\times\left(2+2^2\right)\)
\(=\left(2+2^2\right)\times\left(1+2^2+2^3+...+2^{2008}\right)\)
\(=6\times\left(2^2+2^3+...+2^{2008}\right)\)
\(=3\times2\times\left(2^2+2^3+...+2^{2008}\right)\)
\(\Rightarrow A⋮3\)
*Ta có: A \(=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(=2\times\left(1+2+2^2\right)+2^4\times\left(1+2+2^2\right)+...+2^{2008}\times\left(1+2+2^2\right)\)
\(=\left(1+2+2^2\right)\times\left(2+2^4+2^7+...+2^{2008}\right)\)
\(=7\times\left(2+2^4+2^7+...+2^{2008}\right)\)
\(\Rightarrow A⋮7\)
Mình sửa lại đề C 1 chút xíu
*Ta có: C \(=3^1+3^2+3^3+3^4+...+3^{2010}\)
\(=\left(3+3^2\right)+3^2\times\left(3+3^2\right)+...+3^{2008}\times\left(3+3^2\right)\)
\(=\left(3+3^2\right)\times\left(1+3^2+3^3+...+3^{2008}\right)\)
\(=12\times\left(1+3^2+3^3+...+3^{2008}\right)\)
\(=4\times3\times\left(1+3^2+3^3+...+3^{2008}\right)\)
\(\Rightarrow C⋮4\)
Các câu khác làm tương tự nhé. Chúc bạn học tốt!
Bài 1: a) \(M=1+5+5^2+...+5^{100}\)
\(5M=5+5^2+5^3+...+5^{101}\)
\(5M-M=\left(5+5^2+5^3+...+5^{101}\right)-\left(1+5+5^2+...+5^{100}\right)\)
\(4M=5^{101}-1\)
\(M=\frac{5^{101}-1}{4}\)
b) \(N=2+2^2+...+2^{100}\)
\(2N=2^2+2^3+...+2^{101}\)
\(2N-N=\left(2^2+2^3+...+2^{101}\right)-\left(2+2^2+...+2^{100}\right)\)
\(N=2^{101}-2\)
Bài 2:
a) \(16^{32}=\left(2^4\right)^{32}=2^{128}\)
\(32^{16}=\left(2^5\right)^{16}=2^{80}\)
Vì \(2^{128}>2^{80}\Rightarrow16^{32}>32^{16}\)
a,\(5^3.2-100:4+2^3.5\)
= 125 . 2 - 25 + 8 . 5
= 250 - 25 + 40
= 265
b, \(6^2:9+50.2-3^3.3\)
= 36 : 9 + 100 - 27 . 3
= 4 + 100 - 81
= 23
a) \(A=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(A=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\)
\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)
\(A=3\left(2+2^3+...+2^{2009}\right)⋮3\)
\(A=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(A=\left(2^1+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\)
\(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)\)
\(A=7\left(2^1+2^4+...+2^{2008}\right)⋮7\)
Các ý dưới bạn làm tương tự nhé.
a) \(100-\left(3.5^2-2.3^3\right)\)
\(100-\left(75-52\right)=100-23=77\)
b)\(4.5^2-3.2^3+3^9:3^7\)
\(100-24+9=85\)
c) \(\left(392:7\right).2^3-2^3+2020^0\)
\(8\left(56-1\right)+1=441\)
d) \(\left(6^2+7^2+8^2+9^2+10^2\right)-\left(1^2+2^2+3^2+4^2+5^2\right)=275\)
(lười - thông cảm :(
Đây bạn ơi a,\(100-\left(3.5^2-2.3^3\right)=100-\left(75-54\right)\))=100-19=81
b,\(4.5^2-3.2^3+\frac{3^9}{3^7}=100-24+9=85\)
c,\(\left(392:7\right).2^3-2^3+2020^0=56.2^3-2^3+1=441\)
d,\(=\left(6^2-1^2\right)+\left(7^2-2^2\right)+\left(8^2-3^3\right)+\left(9^2-4^2\right)+\left(10^2-5^2\right)\)
=\(5.7+5.9+5.11+5.13+5.15=5\left(7+9+11+13+15\right)\)
=\(5.55=275\)
(1981 x 1982 - 990) : (1980 x 1982 + 992)
=(1980 x 1982+1982 -990) : (1980 x 1982 +992)
=(1980 x 1982 + 992) : ( 1980 x 1982 + 992)
=1
B=[(45.79+45.21)]:90-5^2]:5+2^3 B=[(45.79+45.21):90-25]:5+8 B=[(45.(79+21):65]:13 B=[(45.100):65]:13 B=[4500:65]:13 B=4500:65:13