Question

1) Cho a, b, c thỏa: abc ≠ 0 và ab + bc + ca = 0. Tính A = (a + b)(b + c)(c + a) /abc

Answer 1

\(A=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\\ =\dfrac{\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\right]-abc}{abc}\\ =\dfrac{\left(abc+a^2b+ac^2+a^2c+b^2c+b^2a+bc^2+abc+abc\right)-abc}{abc}\\ =\dfrac{\left[ab\left(a+b\right)+abc+bc\left(b+c\right)+abc+ca\left(a+c\right)+abc\right]-abc}{abc}\\ =\dfrac{\left[ab\left(a+b+c\right)+bc\left(a+b+c\right)+ca\left(a+b+c\right)\right]-abc}{abc}\\ =\dfrac{\left(ab+bc+ca\right)\left(a+b+c\right)-abc}{abc}\\ =\dfrac{0\cdot\left(a+b+c\right)-abc}{abc}\\ =\dfrac{-abc}{abc}\\ =-1\)