Ta có:
\(\left(a+b\right)\left(b+c\right)\left(c+d\right)\left(d+a\right)\)
\(=\left(\frac{2017}{c}+\frac{2017}{d}\right)\left(\frac{2017}{d}+c\right)\left(c+d\right)\left(d+\frac{2017}{c}\right)\)
\(=\frac{2017}{c^2d^2}\left(c+d\right)^2\left(cd+2017\right)^2\)
\(=\frac{2017}{c^2d^2}\left(c^2d+d^2c+2017c+2017d\right)^2\left(1\right)\)
Ta lại có:
\(\left(a+b+c+d\right)^2\)
\(=\left(\frac{2017}{c}+\frac{2017}{d}+c+d\right)^2\)
\(=\frac{1}{c^2d^2}\left(c^2d+d^2c+2017c+2017d\right)^2\left(2\right)\)
Từ (1) và (2) \(\Rightarrow M=2017\)
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Thay vào tính
tôi ko biết cách giải bài toán này
câu hỏi tao lao vcl neu be ra rồi uống lở uông nham hai vien (đỏ thì sao)---???---
'''''''''''''''''''''hết'''''''''''''''''''
M=(a.c)+(b.d)+(a.c)+(b.d) phan (a.c+b.d)2
M=2017.4 phan (2017+2017)2
M=8068 phan16273156
Vì \(ab=bd=2017\Rightarrow a=b=c=d\) .
\(\Rightarrow\left(ab\right)=\left(bc\right)=\left(cd\right)=\left(ac\right)=\left(ad\right)=\left(bd\right)=\left(ca\right)=\left(ba\right)\)
Theo tính chất tỉ dãy số bằng nhau ta có:
\(M=\frac{\left(a+b\right)\left(b+c\right)\left(c+d\right)\left(d+a\right)}{\left(a+b+c+d\right)^2}\)
\(\Leftrightarrow\frac{\left(ab\right)\left(bc\right)\left(cd\right)\left(ac\right)\left(ad\right)\left(bd\right)\left(ca\right)\left(ba\right)}{\left[\left(ab\right)\left(bc\right)\left(cd\right)\left(ac\right)\left(ad\right)\left(bd\right)\left(ca\right)\left(ba\right)\right]^2}\Leftrightarrow\frac{2017}{\left(2017\right)^2}=\frac{1}{2}\)
\(\Rightarrow M=\frac{1}{2}\)
Dấu = xảy ra khi a = b = c =d
