Cho hình vuông như Hình 1.12. Em hãy thay mỗi dấu “?” bằng một lũy thừa của 2, biết các lũy thừa trên mỗi hàng, mỗi cột và mỗi đường chéo đều bằng nhau.
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tích của mỗi hàng, mỗi cột,mỗi đường chéo là:
27.24.21 = 27+ 4+ 1 = 212
Từ đó ta điền được vào các ô trống còn lại như sau:
27 | 20 | 25 |
22 | 24 | 26 |
23 | 28 | 21 |
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm:
\(2^7\) | \(2^0\) | \(2^5\) |
\(2^2\) | \(2^4\) | \(2^6\) |
\(2^3\) | \(2^8\) | \(2^1\) |
![](https://rs.olm.vn/images/avt/0.png?1311)
Tích của mỗi hàng, cột, đường chéo là:
100.10-5.102 = 10–3
Từ đó ta điền được vào các ô trống còn lại như sau:
100 | 10-5 | 102 |
101 | 10-1 | 10-3 |
10-4 | 103 | 10-2 |
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Tổng các số trong bảng = Tổng các hàng
Mà tổng các hàng bằng 0 nên tổng các số trong bảng đó bằng 0.
b) Xét hàng số 1 ta có:
a + (-2) + (-1) = 0 => a + (-3) = 0 => a = 3
Xét cột số 1 ta có:
3 + (-4) + d = 0 => (-1) + d = 0 => d = 1.
Xét đường chéo chứa b và d có:
(-1) + b + d = 0 => (-1) + b +1 = 0 => b = 0.
Xét cột số 2 ta có:
(-2) + 0 + e = 0 => e = 2
Xét dòng số 2 có:
-4 + b + c = 0 => -4 + 0 + c = 0 => c =4
Xét dòng số 3 có:
d + e + g = 0 => 1 + 2 + g = 0 => g = -3
![](https://rs.olm.vn/images/avt/0.png?1311)
mình cũng có câu hỏi như thế mình cũng học cô loan mà đúng không bà là cẩm thanh mà bà cũng học cô loan mà
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Ta đặt các ô chưa biết như sau:
Ta có:
Tích của mỗi hàng, cột, đường chéo bằng: \({2^3}{.2^4}{.2^5} = {2^{3 + 4 + 5}} = {2^{12}}\)
\(\begin{array}{l}A = {2^{12}}:{2^6}:{2^5} = {2^{12 - 6 - 5}} = {2^1} = 2;\\B = {2^{12}}:{2^1}:{2^3} = {2^{12 - 1 - 3}} = {2^8};\\C = {2^{12}}:{2^8}:{2^4} = {2^{12 - 8 - 4}} = {2^0} = 1;\\D = {2^{12}}:{2^0}:{2^5} = {2^{12 - 0 - 5}} = {2^7};\\E = {2^{12}}:{2^7}:{2^3} = {2^{12 - 7 - 3}} = {2^2}\end{array}\)
Vậy ta có bảng hoàn chỉnh là: