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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.
*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!
Mình xin trả lời:
212 =1025; 73 =343 => 210 < 3.73 => (210)238 <3238 .(73)238 => 22380< 3238 . 7714
28= 256; 34 =243=> 35 < 28
Ta có : 3238= 33. 2225 = 33. (35) 47 < 25. 2376 => 3328 < 2381
22380 < 2238 . 7714 => 21999 < 714 mà 21999> 21993 => 21993< 7714
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Ê bạn vào chỗ https://olm.vn/hoi-dap/question/911743.html
\(2^{10}=1024< 1029=3.7^3\)
\(\Leftrightarrow\left(2^{10}\right)^{238}< \left(3.7^3\right)^{238}\)
\(\Leftrightarrow2^{2380}< 3^{238}.7^{714}\) \(\left(1\right)\)
\(3^5=243< 256=2^8\) \(\left(2\right)\)
\(3^3=27< 32=2^5\) \(\left(3\right)\)
Từ \(\left(2\right)\), \(\left(3\right)\) ta có:
\(3^{328}=3^3.3^{325}=3^3\left(3^5\right)^{47}< 2^5\left(2^8\right)^{47}=2^{381}\)\(\left(4\right)\)
Từ \(\left(1\right)\), \(\left(4\right)\) ta có:
\(2^{2380}< 3^{238}.7^{714}\)
\(\Leftrightarrow2^{2380}< 2^{381}.7^{714}\)
\(\Leftrightarrow2^{1999}< 7^{714}\)
\(\Leftrightarrow2^{1993}< 7^{714}\).
212 = 1025 ; 73 = 343 \(\Rightarrow\) 210 < 3.73 \(\Rightarrow\)\(\left(2^{10}\right)^{238}< 3^{238}.\left(7^3\right)^{238}\)\(\Rightarrow\)22380 < 3238.7714 .
28 = 256 ; 34 = 243 => 35 < 2^8
Ta có : 3328 = 33.2225 = \(3^3.\left(3^5\right)^{47}< 3^3.\left(2^8\right)^{47}< 2^5.2^{376}\Rightarrow3^{328}< 2^{381}\)
22380 < 2238.7714 => 22380 < 2238.7714 => 21999 < 714 mà 21999 > 21993 => 21993 < 7714 .
Đặt A=2+22+23+24+...+22016
- A=(2+22)+(23+24)+...+(22015+22016)
A=2(1+3)+23(1+2)+...22015(1+2)
A=2.3+23.3+...+22015.3
A=3.(2+23+...+22015)chia hết cho 3
A=(2+22+23)+(24+25+26)+...+(22014+22015+22016)
A=2(1+2+22)+24(1+2+22)+...+22014(1+2+22)
A=2.7+24.7+...+22014.7
A=7.(2+24+...+22016)chia hết cho 7
\(A=2+2^2+2^3+...+2^{60}\)
\(=\left(2+2^2+2^3+2^4\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(=2\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)\)
\(=15\left(2+...+2^{57}\right)⋮5\)
\(A=2+2^2+2^3+2^4+...+2^{60}\)
\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
\(=7\left(2+2^4+...+2^{58}\right)⋮7\).
hoàn đức hà là giáo viên trên olm phải ko?
1) (5+54)+(52+55)+...........+(52003+52006)= 5(1+53)+52(1+53)+..............+52003(1+53)
= (5+52+..........+52003).126 ->S chia hết cho 126
2, 7+73+................+71997+71999 = 7(1+72)+..............+71997(1+72)
= (7+...............+71997).50-> chia hết cho 5
= 7(1+72+.......+71998) -> chia hết cho 7
-> chia hết cho 35
Chứng minh cái gì? chắc là so sánh
\(7^{714}< 8^{714}\)
\(2^{1999}>2^{1998}=\left(2^3\right)^{666}=8^{666}\)
\(\Rightarrow2^{1999}>8^{666}>8^{714}>7^{714}\)