Đáp án: $1$
Giải thích các bước giải:
Ta có $abc(a+b+c)=1$
$\to ab.ac+ba.bc+ca.cb=1$
Đặt $ab=x, bc=y, ca=z$
$\to xz+xy+yz=1$
Ta có:
$P=\dfrac{c^2(a+b)^2(a^2b^2+1)}{(b^2c^2+1)(c^2a^2+1)}$
$\to P=\dfrac{(ca+bc)^2((ab)^2+1)}{((bc)^2+1)((ca)^2+1)}$
$\to P=\dfrac{(z+y)^2(x^2+1)}{(y^2+1)(z^2+1)}$
$\to P=\dfrac{(z+y)^2(x^2+xy+yz+zx)}{(y^2+xy+yz+zx)(z^2+xy+yz+zx)}$ vì $xy+yz+zx=1$
$\to P=\dfrac{(z+y)^2(x+y)(x+z)}{(y+x)(y+z)(z+x)(z+y)}$
$\to P=\dfrac{(z+y)^2(x+y)(x+z)}{(z+y)^2(x+y)(x+z)}$
$\to P=1$
