[Toán 9]

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vipboycodon

$\dfrac{1}{a+ab+1}+\dfrac{1}{b+bc+1}+\dfrac{1}{c+ac+1}$

$= \dfrac{1}{a+ab+1}+\dfrac{a}{ab+abc+a}+\dfrac{ab}{abc+a^2bc+ab}$

$= \dfrac{1}{a+ab+1}+\dfrac{a}{a+ab+1}+\dfrac{ab}{a+ab+1}$

$= 1$

Đề có dấu nhỏ hơn hoặc bằng nữa à , kì nhỉ :))
 
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