[Toán 10] Tính giá trị biểu thức

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lucas9999

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V

vansang02121998

$(a+b+c)(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})=1$

$\Leftrightarrow \dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}+2=0$

$\Leftrightarrow a(\dfrac{1}{b}+\dfrac{1}{c})+\dfrac{b+c}{c}+\dfrac{b+c}{b}+\dfrac{b+c}{a}=0$

$\Leftrightarrow \dfrac{a(b+c)}{bc}+\dfrac{(b+c)^2}{bc}+\dfrac{b+c}{a}=0$

$\Leftrightarrow (b+c)(\dfrac{a+b+c}{bc}+\dfrac{1}{a})=0$
 
V

vansang02121998

Rồi tính T tiếp như thế nào hả anh************************************************************************************??

$(b+c)(\dfrac{a+b+c}{bc}+\dfrac{1}{a})=0$

$\Leftrightarrow \dfrac{(b+c)(a+b)(a+c)}{abc}=0$

$\Rightarrow (b+c)(a+b)(a+c)=0$

$\Rightarrow a=-b$ hoặc $b=-c$ hoặc $c=-a$

$\Rightarrow P=0$
 
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