Question

cho \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1\) . chứng minh rằng : \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=0\)

Answer 1

Ta có:

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

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

=>\(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}+c\left(\dfrac{a}{a+b}+\dfrac{b}{a+b}\right)+b\left(\dfrac{a}{a+c}+\dfrac{c}{a+c}\right)+a\left(\dfrac{b}{b+c}+\dfrac{c}{b+c}\right)=a+b+c\)

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

=>\(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}=0\)(đpcm)