Cho a+b+c = 0 Rút gọn A = $\frac{a^2}{bc}$ + $\frac{b^2}{ac}$ + $\frac{c^2}{ab}$ B = $\frac{a^2}{a^2-b^2-c^2}$ + $\frac{b^2}{b^2-c^2-a^2}$ + $\frac{

0 bình luận về “Cho a+b+c = 0 Rút gọn A = $\frac{a^2}{bc}$ + $\frac{b^2}{ac}$ + $\frac{c^2}{ab}$ B = $\frac{a^2}{a^2-b^2-c^2}$ + $\frac{b^2}{b^2-c^2-a^2}$ + $\frac{”

1. Giải thích các bước giải:

   Ta có:

   $a+b+c=0$

   $\to (a+b+c)^3=0$

   $\to a^3+b^3+c^3+3(a+b)(b+c)(c+a)=0$

   $\to a^3+b^3+c^3+3(-c)(-a)(-b)=0$

   $\to a^3+b^3+c^3-3abc=0$

   $\to a^3+b^3+c^3=3abc$

   $\to \dfrac{a^3}{abc}+\dfrac{b^3}{abc}+\dfrac{c^3}{abc}=3$

   $\to \dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}=3$

   $\to A=3$

   Ta có:

   $a+b+c=0$

   $\to b+c=-a$

   $\to (b+c)^2=(-a)^2$

   $\to b^2+c^2+2bc=a^2$

   $\to a^2-b^2-c^2=2bc$

   $\to \dfrac{a^2}{a^2-b^2-c^2}=\dfrac{a^2}{2bc}$

   Tương tự

   $\dfrac{b^2}{b^2-c^2-a^2}=\dfrac{b^2}{2ca}$

   $\dfrac{c^2}{c^2-a^2-b^2}=\dfrac{c^2}{2ab}$

   $\to B=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2ab}$

   $\to B=\dfrac12(\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab})$

   $\to B=\dfrac12A$

   $\to B=\dfrac32$

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