Hình vuông có 3x3 ô chứa 9 số mà tổng các số ở mỗi hàng, mỗi cột và mỗi đường chéo đều = nhau gọi là hình vuông kì diệu. CMR số ở tâm (X) của 1 hình vuông kì diệu = trung bình cộng của hai số còn lại cùng hàng, hoặc cùng cột, hoặc cùng đường chéo
X | ||
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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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kj
a | b | c |
d | e | f |
g | h | i |
Theo đề bài, ta có;
\(a+b+c=a+d+g=c+f+i=g+h+i\)
\(=b+e+h=d+e+f=a+e+i=c+e+g\)
Từ đó ta có \(a+b+c+a+d+g+c+f+i+g+h+i\)\(=b+e+h+d+e+f+a+e+i+c+e+g\)
hay \(2a+2c+2g+2i+b+d+f+h=4e+a+b+c+d+f+g+h+i\)
hay \(a+c+g+i=4e\) (1)
Mặt khác \(a+b+c=b+e+h\)\(\Leftrightarrow a+c=e+h\)
Và \(g+h+i=b+e+h\)\(\Leftrightarrow g+i=b+e\)
Vậy \(4e=e+b+e+h\)hay \(2e=b+h\)hay \(4e=2\left(b+h\right)=\left(b+h\right)+\left(b+h\right)\)
Do \(d+e+f=b+e+h\)nên \(d+f=b+h\), từ đó \(4e=b+d+f+h\)(2)
Từ (1) và (2) ta có: \(8e=a+b+c+d+f+g+h+i\)hay \(e=\frac{a+b+c+d+f+g+h+i}{8}\)
Và đó là đpcm
http://www.olm.vn/hoi-dap/question/95083.html
có người làm rùi mà