Bạn chưa đăng nhập. Vui lòng đăng nhập để hỏi bài

Những câu hỏi liên quan
PHỞ ĐẶC BIỆT
Xem chi tiết
Hieu Mai
27 tháng 3 2018 lúc 21:48

Áp dụng hằng đẳng thức sau
an−1=(a−1).[an−1+an−2+...+1]=(a−1).pan−1=(a−1).[an−1+an−2+...+1]=(a−1).p (nn là 1 số nguyên dương)
an+1=(a+1).[an−1−an−2+..+1]=(a+1).qan+1=(a+1).[an−1−an−2+..+1]=(a+1).q (nn là 1 số nguyên dương lẻ)

Thay vào ta được như sau:

+) 222333−1=(222−1).p=13.17.p222333−1=(222−1).p=13.17.p

+) 333222+1=(3332)111+1=110889111+1=(110889+1).q=13.8530.q333222+1=(3332)111+1=110889111+1=(110889+1).q=13.8530.q

=>=> 222333+333222=222333−1+333222+1=13(17p+8530q)⋮13222333+333222=222333−1+333222+1=13(17p+8530q)⋮13

Vậy: 222333+333222⋮13222333+333222⋮13 (đpcm)(đpcm) 

Lê Hoàng Bảo Châu
Xem chi tiết
Yuzuri Yukari
20 tháng 9 2016 lúc 12:41

\(\left(222^{333}+333^{222}\right)⋮13\)

an−1=(a−1).[an−1+an−2+...+1]=(a−1).p" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvetica,arial,sans-serif; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">n" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvetica,arial,sans-serif; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">an+1=(a+1).[an−1−an−2+..+1]=(a+1).q" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvetica,arial,sans-serif; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">n" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvetica,arial,sans-serif; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">222333−1=(222−1).p=13.17.p" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvetica,arial,sans-serif; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">
333222+1=(3332)111+1=110889111+1=(110889+1).q=13.8530.q" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvetica,arial,sans-serif; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">
=>" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvetica,arial,sans-serif; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">

Nguyễn Nam
2 tháng 12 2017 lúc 22:10

a) \(222^{333}+333^{222}\)

\(=\left(111.2\right)^{333}+\left(111.3\right)^{222}\)

\(=111^{333}.2^{333}+111^{222}.3^{222}\)

\(=111^{222}.\left(111^{111}.2^{333}+3^{222}\right)\)

\(=111^{222}.\left(111^{111}.2^{3.111}+3^{2.111}\right)\)

\(=111^{222}.\left[111^{111}.\left(2^3\right)^{111}+\left(3^2\right)^{111}\right]\)

\(=111^{222}.\left(111^{111}.8^{111}+9^{111}\right)\)

\(=111^{222}.\left[\left(111.8\right)^{111}+9^{111}\right]\)

\(=111^{222}.\left(888^{111}+9^{111}\right)\)

\(=111^{222}.\left(888+9\right)\left[888^{110}-888^{109}.9+.....-888.9^{109}+9^{110}\right]\)

\(=111^{222}.7992\left[888^{110}-888^{109}.9+.....-888.9^{109}+9^{110}\right]\)

\(=111^{222}.897\left[888^{110}-888^{109}.9+.....-888.9^{109}+9^{110}\right]\)

\(=111^{222}.13.69\left[888^{110}-888^{109}.9+.....-888.9^{109}+9^{110}\right]⋮13\)

Vậy \(222^{333}+333^{222}⋮13\left(dpcm\right)\)

Nguyễn Minh Hoàng
Xem chi tiết
tthnew
18 tháng 1 2021 lúc 19:41

a)

Ta có: \(222^{333}=\left(222^3\right)^{111}\equiv1^{111}=1\left(mod13\right)\)

\(\Rightarrow222^{333}+333^{222}\equiv1+333^{222}=1+\left(333^2\right)^{111}\)

\(\equiv1+12^{111}\equiv1+12^{110}\cdot12\equiv1+\left(12^2\right)^{55}\cdot12\)

\(\equiv1+1\cdot12\equiv13\equiv0\left(mod13\right)\)

Vậy $222^{333}+333^{222}$ chia hết cho $13.$

b) Ta có:

\(3^{105}\equiv\left(3^3\right)^{35}\equiv1^{35}\equiv1\) (mod13)

\(\Rightarrow3^{105}+4^{105}\equiv1+4^{105}\equiv1+\left(4^3\right)^{35}\)

\(\equiv1+12^{35}\equiv1+\left(12^2\right)^{17}\cdot12\equiv1+1\cdot12\equiv13\equiv0\left(mod13\right)\)

Vậy $3^{105}+4^{105}$ chia hết cho $13.$

Lại có:

\(3^{105}\equiv\left(3^3\right)^{35}\equiv5^{35}\equiv\left(5^5\right)^7\equiv1\left(mod11\right)\)

\(4^{105}\equiv\left(4^3\right)^{35}\equiv9^{35}\equiv\left(9^5\right)^7\equiv1\left(mod11\right)\)

Từ đây:\(3^{105}+4^{105}\equiv1+1\equiv2\left(mod11\right)\)

Vậy $3^{105}+4^{105}$ không chia hết cho $11.$

P/s: Rất lâu rồi không giải, không chắc.

Nguyễn Minh Hoàng
Xem chi tiết
Nguyễn Thị Huyền Trang
Xem chi tiết
Mac Phuong Nga
Xem chi tiết
Nguyễn Đặng Thảo Nhi
Xem chi tiết
Nguyễn Lê Anh Quân
Xem chi tiết
phung thi hien
Xem chi tiết
than duc minh
10 tháng 3 2016 lúc 22:04

du 2 h cho minh nha