Đáp án:
Ta có
`1/a + 1/b + 1/c = 0`
`<=> (ab + bc + ca)/(abc) = 0`
`<=> ab + bc + ca = 0`
`<=> {ab = -(bc + ca)`
`{bc = - (ab + ca)`
`{ca = - (ab + bc)`
`+) a^2 + 2bc = a^2 + bc - ab - ca = a(a - b) - c(a - b) = (a - b)(a - c)`
`-> (bc)/(a^2 + 2bc) = = (bc)/[(a - b)(a - c)] = (bc(b - c))/[(a - b)(a - c)(b - c)]`
tương tự
`-> (ac)/(b^2 + 2ac) = (ac)/[(b - a)(b - c)] = [-ac(a - c)]/[(a - b)(a - c)(b - c)]`
`(ab)/(c^2 + 2ab) = (ab)/[(c - a)(c - b)] = [ab(a - b)]/[(a - b)(a - c)(b - c)]`
`-> M = (bc)/(a^2 + 2bc) + (ac)/(b^2 + 2ac) + (ab)/(c^2 + 2ab)`
`= (bc(b - c))/[(a - b)(a - c)(b - c)] + [-ac(a - c)]/[(a - b)(a - c)(b - c)] + [ab(a - b)]/[(a - b)(a - c)(b - c)]`
`= [bc(b - c) - ac(a - c) + ab(a - b)]/[(a - b)(a - c)(b - c)]`
`= [(a - b)(a - c)(b - c)]/[(a - b)(a - c)(b - c)]`
`= 1`
____________
`bc(b - c) - ac(a - c) + ab(a - b)`
`= bc(b - c) - a^2c + ac^2 + a^2b - ab^2`
`= bc(b - c) + a^2(b - c) - a(b^2 - c^2)`
`= bc(b - c) + a^2(b - c) - a(b - c)(b + c)`
`= (b - c)(bc + a^2 - ab - ac)`
`= (b - c)[a(a - c) - b(a - c)]`
`= (a - b)(b - c)(a - c)`
Giải thích các bước giải:
