Question

[phan gia huy]

Cho ac = bd = 2017 , (a, b, c, d > 0 )

Tính \(M=\frac{\left(a+b\right)\left(b+c\right)\left(c+d\right)\left(d+a\right)}{\left(a+b+c+d\right)^2}\)

Answer 1

[alibaba nguyễn]

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) \(\Rightarrow M=2017\)

Answer 2

[Bexiu]

LỜI GIẢI 

a+cb+d=a−cb−da+cb+d=a−cb−d

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

a+cb+d=a−cb−d=a+c+a−cb+d+b−d=2a2b=ab(1)a+cb+d=a−cb−d=a+c+a−cb+d+b−d=2a2b=ab(1)

a+cb+d=a−cb−d=a+c−a+cb+d−b+d=2c2d=cd(1)a+cb+d=a−cb−d=a+c−a+cb+d−b+d=2c2d=cd(1)

Từ (1)(1) và (2)(2) ta có:

ab=cdab=cd

Đặt:

ab=cd=kab=cd=k ⇒{a=bkc=dk⇒{a=bkc=dk

Thay vào tính