Question

[Nguyen Thi Ngoc Linh]

Bai 1: Tinh

A = 253 x 75 - 161 x 37 + 253 x 25 - 161 x 63 / 100 x 47 - 12 x 3,5 - 5,8 : 0,1

Bai 2: Tim quy luat va tinh tong cua moi day:

a,1;3;9;27;...;729

b,1;2;4;8;...;2048

c,1;4;9;16;...;1089

Minh se tich diem cho cac ban ! giai ra nhe cac ban!

Answer 1

[Mới vô]

1.

\(A=\dfrac{253\cdot75-161\cdot37+253\cdot25-161\cdot63}{100\cdot47-12\cdot3,5-5,8:0,1}\\ =\dfrac{253\cdot75+253\cdot25-\left(161\cdot37+161\cdot63\right)}{100\cdot47-\left(12\cdot3,5+5,8:0,1\right)}\\ =\dfrac{253\cdot\left(75+25\right)-161\cdot\left(37+63\right)}{100\cdot47-\left(42+58\right)}\\ =\dfrac{253\cdot100-161\cdot100}{100\cdot47-100}\\ =\dfrac{100\cdot\left(253-161\right)}{100\cdot\left(47-1\right)}\\ =\dfrac{253-161}{47-1}\\ =\dfrac{92}{46}\\ =2\)

Answer 2

[Mới vô]

2.

a,

Quy luật: Số thứ \(n=3^{n-1}\left(1\le n\le6\right)\)

Dãy số tương đương với:

\(3^{1-1};3^{2-1};3^{3-1};...;3^{7-1}\\ \Leftrightarrow3^0;3^1;3^2;...;3^6\)

\(3^0+3^1+3^2+...+3^6\\ =1\cdot\left(3^0+3^1+3^2+...+3^6\right)\\ =\dfrac{3-1}{2}\cdot\left(3^0+3^1+3^2+...+3^6\right)\\ =\dfrac{\left(3-1\right)\cdot\left(3^0+3^1+3^2+...+3^6\right)}{2}\\ =\dfrac{3^1-3^0+3^2-3^1+3^3-3^2+...+3^7-3^6}{2}\\ =\dfrac{3^7-1}{2}\\ =\dfrac{2187-1}{2}\\ =\dfrac{2186}{2}\\ =1093\)

b,

Quy luật: Số thứ \(n=2^{n-1}\left(1\le n\le12\right)\)

Dãy số tương đương với:

\(2^{1-1};2^{2-1};2^{3-1};...;2^{12-1}\\ \Leftrightarrow2^0;2^1;2^2;...;2^{11}\)

\(2^0+2^1+2^2+...+2^{11}\\ =1\cdot\left(2^0+2^1+2^2+...+2^{11}\right)\\ =\left(2-1\right)\cdot\left(2^0+2^1+2^2+...+2^{11}\right)\\ =2^1-2^0+2^2-2^1+2^3-2^2+...+2^{12}-2^{11}\\ =2^{12}-1\\ =4096-1\\ =4095\)

c,

Quy luật: Số thứ \(n=n^2\left(1\le n\le33\right)\)

Dãy số tương đương với:

\(1^2;2^2;3^2;...;33^2\)

\(1^2+2^2+3^2+...+33^2\\ =\dfrac{33\cdot34\cdot65}{6}\\ =12155\)

Cách chứng minh: https://diendan.hocmai.vn/threads/1-2-2-2-n-2.220374/