a) Cho a + b + c = 0. Đặt P = $\frac{a-c}{c}$ +$\frac{b-c}{a}$ +$\frac{c-a}{b}$ ; Q = $\frac{c}{a-b}$ +$\frac{a}{b-c}$+ $\frac{b}{c-a}$ Tính P x Q

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1. Giải thích các bước giải:

   Đặt A = P x Q

   $A=\left ( \frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b} \right ).\left ( \frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a} \right )$

    

   Ta có: $\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}=\frac{ab(a-b)+bc(b-c)+ca(c-a)}{abc} =\frac{b(a^2-ab+cb-c^2)+ca(c-a)}{abc}=\frac{b\left [ (a-c)(a+c)-b(a-c)+ca(c-a) \right ]}{abc}$

    

   $=\frac{b(a-c)(a+c-b)+ca(c-a)}{abc}=\frac{(a-c)(ba+bc-b^2-ca)}{abc} =\frac{(a-c)\left [ a(b-c)-b(b-c) \right ]}{abc}=\frac{(a-c)(b-c)(a-b)}{abc}$

    

   Lại có: $\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}=\frac{c(b-c)(c-a)+a(a-b)(c-a)+b(a-b)(b-c)}{(a-b)(b-c)(c-a)}$

    

   $=\frac{(c-a)(cb-c^2+a^2-ab)+b(a-b)(b-c)}{(a-b)(b-c)(c-a)}=\frac{(c-a)(b(c-a)-(c-a)(c+a))+b(a-b)(b-c)}{(a-b)(b-c)(c-a)}$

    

   $=\frac{(c-a)^2(b-c-a)+b(a-b)(b-c)}{(a-b)(b-c)(c-a)}=\frac{2b.(c-a)^2+b(a-b)(b-c)}{(a-b)(b-c)(c-a)} (vì c+a=-b)$

    

   $=\frac{b\left [ 2(c-a)^2+(a-b)(b-c) \right ]}{(a-b)(b-c)(c-a)}=\frac{b\left [ 2(c-a)^2+ab-ac-b^2+bc \right ]}{(a-b)(b-c)(c-a)}$

    

   $=\frac{b\left [ 2(c-a)^2+b(c+a)-ac-b^2 \right ]}{(a-b)(b-c)(c-a)}=\frac{b\left [ 2(c-a)^2+b.(-b)-ac-b^2 \right ]}{(a-b)(b-c)(c-a)}$

    

   $=\frac{b\left [ 2(c^2+a^2)-5ac-2b^2 \right ]}{(a-b)(b-c)(c-a)}=\frac{b\left [ 2(c^2+a^2)-5ac-2a^2-2c^2-4ac \right ]}{(a-b)(b-c)(c-a)} =\frac{-9abc}{(a-b)(b-c)(c-a)} (vì -b=c+a\Rightarrow b^2=c^2+a^2+2ac\Leftrightarrow 2b^2=2a^2+2c^2+4ac)$

    

   Vậy $A=\left ( \frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b} \right ).\left ( \frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a} \right ) =\frac{(a-c)(b-c)(a-b)}{abc}.\frac{-9abc}{(a-b)(b-c)(c-a)}=9$

    

    

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