Cho a,b,c ∈R và $(\frac{1}{a}+$ $\frac{1}{b}+$ $\frac{1}{c}=0$ Tính P= $\frac{ab}{c^2}+$ $\frac{bc}{a^2}+$ $\frac{ac}{b^2}$

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1. Có $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0$

   $→\dfrac{1}{a}+\dfrac{1}{b}=-\dfrac{1}{c}$

   $→(\dfrac{1}{a}+\dfrac{1}{b})^3=(-\dfrac{1}{c})^3$

   $→\dfrac{1}{a^3}+\dfrac{1}{b^3}+3.\dfrac{1}{a}.\dfrac{1}{b}.(\dfrac{1}{a}+\dfrac{1}{b})=-\dfrac{1}{c^3}$

   $→\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=-3.\dfrac{1}{a}.\dfrac{1}{b}.(\dfrac{1}{a}+\dfrac{1}{b})$

   Mà $\dfrac{1}{a}+\dfrac{1}{b}=-\dfrac{1}{c}$

   $→\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=-3.\dfrac{1}{a}.\dfrac{1}{b}.-\dfrac{1}{c}$

   $→\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=3.\dfrac{1}{a}.\dfrac{1}{b}.\dfrac{1}{c}=3.\dfrac{1}{abc}$

   Có: $P=\dfrac{ab}{c^2}+\dfrac{bc}{a^2}+\dfrac{ac}{b^2}$

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

   $=abc.(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3})$

   Mà $\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=3.\dfrac{1}{abc}$

   $=abc.3.\dfrac{1}{abc}$

   $=3$

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2. ` có : 1/a +1/b +1/c =0 => 1/a +1/b = -1/c => (1/a +1/b )^3 = (-1/c)^3`
   `=> 1/a^3 + 1/b^3 +3.1/a.1/b . (1/a+1/b) = -1/c^3`
   `=> 1/a^3 +1/b^3 +1/c^3 = -3.1/a.1/b. (1/a+1/b)`
   `Vì 1/a +1/b = -1/c => -3.1/a.1/b .(1/a+1/b)=-3.1/a.1/b.(-1/c)= 3.1/abc`
   `=> 1/a^3 + 1/b^3 +1/c^3 = 3.1/abc`
   `=> abc/a^3 + abc/b^3 + abc/c^3 = 3.abc/abc`
   `=> bc/a^2 +ac/b^2 + ab/c^2 = 3`

   `Nocoyp `

   `Xin câu trả lời hy anhaats`

   Trả lời