Đáp án:

Ta có: `1/a + 1/b + 1/c = 0`

`=> (bc+ac+ab)/(abc) = 0`

`=> ab+bc+ca = 0`

Từ `ab+bc+ca = 0`, ta có:

`bc = -ab-ac`

`ac = -ab-bc`

`ab = -bc-ca`

Suy ra:

`a^2+2bc = a^2+bc+bc = a^2+bc+(-ab-ac) = a(a-b)-c(a-b) = (a-b)(a-c)`

`b^2+2ac = b^2+ac+ac = b^2+ac+(-ab-bc) = b(b-a)-c(b-a) = (b-a)(b-c)`

`c^2+2ab = c^2+ab+ab = c^2+ab+(-bc-ca) = c(c-b)-a(c-b) = (c-a)(c-b)`

a)

`M = 1/(a^2+2bc) + 1/(b^2+2ac) + 1/(c^2+2ab)`

`M = 1/((a-b)(a-c)) + 1/((b-a)(b-c)) + 1/((c-a)(c-b))`

`M = 1/((a-b)(a-c)) - 1/((a-b)(b-c)) + 1/((a-c)(b-c))`

`M = (b-c-a+c+a-b)/((a-b)(a-c)(b-c))`

`M = 0/((a-b)(a-c)(b-c)) = 0`

b)

`N = (bc)/(a^2+2bc) + (ca)/(b^2+2ac) + (ab)/(c^2+2ab)`

`N = (bc)/((a-b)(a-c)) + (ca)/((b-a)(b-c)) + (ab)/((c-a)(c-b))`

`N = (bc)/((a-b)(a-c)) - (ca)/((a-b)(b-c)) + (ab)/((a-c)(b-c))`

`N = (bc(b-c)-ca(a-c)+ab(a-b))/((a-b)(a-c)(b-c))`

Tử `= bc(b-c)-ca(a-c)+ab(a-b)`

`= b^2c-bc^2-a^2c+ac^2+a^2b-ab^2`

`= a^2(b-c)-a(b^2-c^2)+bc(b-c)`

`= (b-c)(a^2-a(b+c)+bc)`

`= (b-c)(a-b)(a-c)`

`N = ((a-b)(b-c)(a-c))/((a-b)(a-c)(b-c)) = 1`

c)

`P = a^2/(a^2+2bc) + b^2/(b^2+2ac) + c^2/(c^2+2ab)`

`P = a^2/((a-b)(a-c)) + b^2/((b-a)(b-c)) + c^2/((c-a)(c-b))`

`P = a^2/((a-b)(a-c)) - b^2/((a-b)(b-c)) + c^2/((a-c)(b-c))`

`P = (a^2(b-c)-b^2(a-c)+c^2(a-b))/((a-b)(a-c)(b-c))`

Tử `=a^2(b-c)-b^2(a-c)+c^2(a-b)`

`= a^2b-a^2c-ab^2+b^2c+ac^2-bc^2`

`= (a-b)(ab-c(a+b)+c^2)`

`= (a-b)(ab-ac-bc+c^2)`

`= (a-b)(a(b-c)-c(b-c))`

`= (a-b)(b-c)(a-c)`

`P = ((a-b)(b-c)(a-c))/((a-b)(a-c)(b-c)) = 1`