Đáp án:
Giải thích các bước giải:
Ta có :
$ \sqrt[3]{\dfrac{NB}{BC}.\dfrac{NF}{AH}} = \sqrt[3]{\dfrac{NB}{BC}.\dfrac{BH}{BC}} = \sqrt[3]{\dfrac{NB.BH}{BC²}}$
$ = \sqrt[3]{(\dfrac{BF}{BC}})² = \sqrt[3]{(\dfrac{BF}{BH}.\dfrac{BH}{BC}})² = \sqrt[3]{(\dfrac{AB}{BC}.\dfrac{BH}{BC}})² $
$ = \sqrt[3]{\dfrac{AB².BH²}{BC^{4}}} = \sqrt[3]{\dfrac{(BH.BC).BH²}{BC^{4}}} = \sqrt[3]{\dfrac{BH³}{BC³}} = \dfrac{BH}{BC} (1)$
Tương tự : $\sqrt[3]{\dfrac{MC}{BC}.\dfrac{ME}{AH}} = \dfrac{CH}{BC} (2)$
$ (1) + (2): \sqrt[3]{\dfrac{NB}{BC}.\dfrac{NF}{AH}} + \sqrt[3]{\dfrac{MC}{BC}.\dfrac{ME}{AH}} =\dfrac{BH}{BC} + \dfrac{CH}{BC} = 1$
$⇔ \sqrt[3]{NB.NF} + \sqrt[3]{MC.ME} = \sqrt[3]{AH.BC}$
$⇔ \sqrt[3]{NB.NF} + \sqrt[3]{MC.ME} = \sqrt[3]{AB.AC}$ ( vì $: AB.AC = AH.BC = 2S_{ABC}$)
