Question

cho a b c khác nhau khác 0 thỏa mãn đk a/b+c=b/c+a=c/a+b

tính gt bt b+c/a=a+c/b=a+b/c

Answer 1

Ta có:\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{b+c+c+a+a+b}=\frac{1}{2}\)

Xét a+b+c=0

\(\Rightarrow\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=\frac{-a}{a}+\frac{-b}{b}+\frac{-c}{c}=-1+-1+-1=-3\)

Xét \(a+b+c\ne0\)

\(\Rightarrow2a=b+c,2b=c+a,2c=a+b\)

\(\Rightarrow\frac{b+c}{a}=2,\frac{a+c}{b}=2,\frac{a+b}{c}=2\Rightarrow\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=2+2+2=6\)

Answer 2

a/b+c =b/c+a =c/a+b
<=> a/b+c +1 = b/c+a +1 = c/a+b +1
<=> a+b+c / b+c = a+b+c / c+a =a+b+c / a+b
<=>a+b+c=0 hoặc b+c= c+a=a+b
nếu b+c=c+a=a+b => a=b=c (vô lý trái với đề bài a, b, c khác nhau)
=> a+b+c=0 => a= - b - c => b+c/a = - 1
tương tự a+b/c = -1 ; a+c/b = - 1
=> P = - 3

Answer 3

Từ giả thiết suy ra :\(ab+bc+ca=0\)
Ta có : \(\frac{\left(b+c\right)}{a}=\frac{\left(ab+ac\right)}{a^2}=\frac{\left(-bc\right)}{a^2}=\frac{\left(-abc\right)}{a^3}\)
Tương tự : \(\frac{\left(a+b\right)}{b}=\frac{\left(-abc\right)}{b^3}\)và \(\frac{\left(a+b\right)}{c}=\frac{\left(-abc\right)}{c^3}\)
Suy ra : M = \(-abc.\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)

Mặt khác : \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
=> \(\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)
=> \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{3.1}{ab}.\left(\frac{1}{a}+\frac{1}{b}\right)=-\frac{1}{c^3}\)
=> \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{3.1}{ab}\left(-\frac{1}{c}\right)=0\)
=> \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{a}{\left(abc\right)}\)

Suy ra : M = \(-\frac{abc.3}{\left(abc\right)}=-3\)