Question

Cho \(\dfrac{ }{ }\)a/b+c +b/a+c + c/a+b =1

Tính A= a²/b+c + b²/a +c +c² /a+b

Answer 1

GT: \(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=1\)

Ta có:

\(A=\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\)

\(A=a\left(\dfrac{a}{b+c}\right)+b\left(\dfrac{b}{a+c}\right)+c\left(\dfrac{c}{a+b}\right)\)

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

\(A=a.\dfrac{a+b+c}{b+c}-a+b.\dfrac{a+b+c}{a+c}-b+c.\dfrac{a+b+c}{a+b}-c\)

\(A=\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\right)-\left(a+b+c\right)\)

\(A=\left(a+b+c\right).1-\left(a+b+c\right)\)

\(A=0\)