Question

Cho abc=24

Tính A=\(\dfrac{3a}{ab+3a+6}+\dfrac{4b}{bc+4b+12}+\dfrac{2c}{ac+2c +8}\)

Answer 1

Ta có

\(\left\{{}\begin{matrix}\dfrac{3a}{ab+3a+6}=\dfrac{3ac}{abc+3ac+6c}=\dfrac{3ac}{24+3ac+6c}=\dfrac{ac}{8+ac+2c}\\\dfrac{4b}{bc+4b+12}=\dfrac{4ab}{abc+4ab+12a}=\dfrac{4ab}{24+4ab+12a}=\dfrac{ab}{6+ab+3a}=\dfrac{abc}{6c+abc+3ac}=\dfrac{24}{6c+24+3ac}=\dfrac{8}{2c+8+ac}\\\dfrac{2c}{ac+2c+8}\end{matrix}\right.\)

=> \(\dfrac{ac}{ac+2c+8}+\dfrac{2c}{ac+2c+8}+\dfrac{8}{ac+2c+8}=\dfrac{ac+2c+8}{ac+2c+8}=1\)

=>A=1