Question

cho ba số a,b,c khác 0 thỏa mãn \(\dfrac{a}{c+b}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=1\)

tính giá trị biểu thức \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\)

Answer 1

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

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

=>\(\dfrac{a^2}{c+b}+\dfrac{ab}{a+c}+\dfrac{ac}{a+b}+\dfrac{b^2}{a+c}+\dfrac{ba}{c+d}+\dfrac{bc}{a+b}+\dfrac{ca}{c+d}+\dfrac{cb}{a+c}+\dfrac{c^2}{a+b}=a+b+c\)=>\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{b\left(a+c\right)}{a+c}+\dfrac{c\left(a+b\right)}{a+b}+\dfrac{a\left(b+c\right)}{c+b}=a+b+c\)=>\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+a+b+c=a+b+c\)

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

chúc bạn học tốt ^ ^