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Lời giải:
Áp dụng TCDTSBN:
$\frac{a+b+c-d}{d}=\frac{b+c+d-a}{a}=\frac{c+d+a-b}{b}=\frac{d+a+b-c}{c}$
$=\frac{a+b+c-d+b+c+d-a+c+d+a-b+d+a+b-c}{d+a+b+c}$
$=\frac{2(a+b+c+d)}{a+b+c+d}=2$
$\Rightarrow a+b+c-d=2d; b+c+d-a=2a; c+d+a-b=2b; d+a+b-c=2c$
$\Rightarrow a+b+c=3d; b+c+d=3a; c+d+a=3b; d+a+b=3c$
Khi đó:
\(P=\frac{a+b+c}{a}.\frac{b+c+d}{b}.\frac{c+d+a}{c}.\frac{a+b+d}{d}\\ =\frac{3d}{a}.\frac{3a}{b}.\frac{3b}{c}.\frac{3c}{d}=81\)
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Ta có : \(a+b+c\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\a+c=-b\end{cases}\left(\cdot\right)}\)
\(\left(1+\frac{a}{b}\right).\left(1+\frac{b}{c}\right).\left(1+\frac{c}{a}\right)\)
\(=\frac{b+a}{b}.\frac{c+b}{c}.\frac{a+c}{a}\)
\(=\frac{-c}{b}.\frac{-a}{c}.\frac{-b}{a}\left(do\cdot\right)\)
\(=-1.-1.-1\)
\(=-1\)
Phải là 1+c/a chứ bạn ơi
A = a+b/b . b+c/c . c+a/a = (-c/b).(-a/c).(-b/a) = -1
k mk nha
Ta có \(a+b+c=0\)
\(\Leftrightarrow\hept{\begin{cases}b+c=-a\left(1\right)\\c+a=-b\left(2\right)\\a+b=-c\left(3\right)\end{cases}}\)
Mặt khác\(A=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{c+a}{a}\left(4\right)\)
Từ\(\left(1\right),\left(2\right),\left(3\right),\left(4\right)\)ta có :
\(A=-\frac{c}{b}\cdot-\frac{a}{b}\cdot-\frac{b}{a}\)
\(=-\frac{a\cdot b\cdot c}{a\cdot b\cdot c}=-1\)
Vậy\(A=-1\)