Question

Rút gọn:

\(B=\left(1+\dfrac{b^2+c^2-a^2}{2bc}\right)\cdot\dfrac{1+\dfrac{a}{b+c}}{1-\dfrac{a}{b+c}}\cdot\dfrac{b^2+c^2-\left(b-c\right)^2}{a+b+c}\)

Answer 1

Ta có: \(B=[\left(b+c\right)^2-a^2].\)\(\dfrac{\dfrac{a+b+c}{b+c}}{\dfrac{b+c-a}{b+c}}\)\(.\)\(\dfrac{2bc}{a+b+c}\)

\(=\left(b+c+a\right)\left(b+c-a\right).\dfrac{a+b+c}{b+c-a}.\dfrac{2bc}{a+b+c}\)

\(=\left(a+b+c\right).2bc=2abc+2b^2c+2bc^2\)