Question

Cho\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\)Tính giá trị của biểu thức\(Q=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)

Answer 1

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

=>\(\frac{a}{b+c}+1=\frac{b}{c+a}+1=\frac{c}{a+b}+1\)

=>\(\frac{a+b+c}{b+c}=\frac{a+b+c}{c+a}=\frac{a+b+c}{a+b}\)

TH1:a+b+c=0

=>\(\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\)

Khi đó Q=\(\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)=\(\frac{a+b}{b}\times\frac{b+c}{c}\times\frac{c+a}{a}\)=\(\frac{-c}{b}\times\frac{-a}{c}\times\frac{-b}{a}\)=-1

TH2: a+b+c\(\ne\)0

=>a=b=c

Vậy Q\(\in\left\{-1;8\right\}\)

Answer 2

đổi 2 dòng cuối thành:

Khi đó Q=(1+1)\(\times\left(1+1\right)\times\left(1+1\right)=2\times2\times2=8\)

Vây Q\(\in\left\{-1;8\right\}\)