Question

cho a,b,c là số thực a,b,c khác nhau a+b+c khác 0 biết a+b-c/c=b+c-a/a=c+a-b/b tình B=(1+b/a)*(1+a/c)*(1+c/b)

Answer 1

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

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

\(=\frac{\left(a+b-c\right)+\left(b+c-a\right)+\left(c+a-b\right)}{c+a+b}=\frac{a+b+c}{a+b+c}=1\)

\(\Rightarrow\hept{\begin{cases}a+b-c=c\\b+c-a=a\\c+a-b=b\end{cases}}\Rightarrow\hept{\begin{cases}a+b+c=3a\\a+b+c=3b\\a+b+c=3c\end{cases}}\Rightarrow a=b=c\)

Khi đó: \(B=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=2^3=8\)

Vậy B = 8

Answer 2

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

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

\(\Leftrightarrow\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{c+a+b}=2\)

\(B=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)\)

\(B=\frac{a+b}{a}.\frac{c+a}{c}.\frac{b+c}{b}\)

\(B=\frac{a+b}{c}.\frac{c+a}{b}.\frac{b+c}{a}=2.2.2=8\)