Question

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

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

Aem nhớ giải rõ ra cho dễ hiểu nhé! Nhanh nhá tui cần gấp lắm

Answer 1

Ta có:

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

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

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

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

\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{\left(a+b\right)+\left(b+c\right)+\left(c+a\right)}{c+a+b}=\frac{2.\left(a+b+c\right)}{a+b+c}=2\)

Ta có: \(B=\left(1+\frac{b}{a}\right).\left(1+\frac{a}{c}\right).\left(1+\frac{c}{b}\right)\)

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

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