`\text{Với điều kiện đề bài đã cho ta có:}`
`\frac{a^2-bc}{a(1-bc)}=\frac{b^2-ac}{b(1-ac)}`
`<=>(a^2-bc)(b-abc)=(b^2-ac)(a-abc)`
`<=>a^2b-a^3bc-b^2c+ab^2c^2=ab^2-ab^2-ab^3c-a^2c+a^bc^2`
`<=>a^2b-ab^2-b^2c+a^c=a^3bc-ab^3c+a^2bc^2-ab^2c^2`
`<=>ab(a-b)+c(a^2-b^2)=abc[(a^2-b^2)+ac-bc]`
`<=>(a-b)(ab+ac+bc)abc(a-b)(a+b+c)`
`<=>ab+ac+bc=abc(a+b+c)`(vì `a ne b=>a-b ne 0)`
`<=>a+b+c=1/a+1/b+1/c`(`do` `abc ne 0)`
