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

Question

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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3 tuần 2021-07-12T10:17:38+00:00 2 Answers 0 views 0

## Answers ( )

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

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

$=3$

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

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