Đáp án:
Giải thích các bước giải:
a)Theo hệ thức lượng trong tam giác vuông:
$ BF² = NB.BH ⇔ NB = \frac{BF²}{BH}$
$ CE² = MC.CH ⇔ MC = \frac{CE²}{CH}$
$ BH.BC = AB²; CH.BC = AC²$
Ta có $: \sqrt[]{MC} + \sqrt[]{NB} = \sqrt[]{\frac{CE²}{CH}} + \sqrt[]{\frac{BF²}{BH}} $
$ = \frac{CE}{\sqrt[]{CH}} + \frac{BF}{\sqrt[]{BH}} = \sqrt[]{BC}(\frac{CE}{\sqrt[]{CH.BC}} + \frac{BF}{\sqrt[]{BH.BC}})$
$ = \sqrt[]{BC}(\frac{CE}{\sqrt[]{AC²}} + \frac{BF}{\sqrt[]{AB²}} = \sqrt[]{BC}(\frac{CE}{AC} + \frac{BF}{AB})$
$ = \sqrt[]{BC}(\frac{CH}{BC} + \frac{BH}{BC}) = \sqrt[]{BC}\frac{CH + BH}{BC} = \sqrt[]{BC}$
b) Ta có :
$ \sqrt[3]{\frac{NB}{BC}.\frac{NF}{AH}} = \sqrt[3]{\frac{NB}{BC}.\frac{BH}{BC}} = \sqrt[3]{\frac{NB.BH}{BC²}}$
$ = \sqrt[3]{(\frac{BF}{BC}})² = \sqrt[3]{(\frac{BF}{BH}.\frac{BH}{BC}})² = \sqrt[3]{(\frac{AB}{BC}.\frac{BH}{BC}})² $
$ = \sqrt[3]{\frac{AB².BH²}{BC^{4}}} = \sqrt[3]{\frac{(BH.BC).BH²}{BC^{4}}} = \sqrt[3]{\frac{BH³}{BC³}} = \frac{BH}{BC} (1)$
Tương tự : $\sqrt[3]{\frac{MC}{BC}.\frac{ME}{AH}} = \frac{CH}{BC} (2)$
$ (1) + (2): \sqrt[3]{\frac{NB}{BC}.\frac{NF}{AH}} + \sqrt[3]{\frac{MC}{BC}.\frac{ME}{AH}} =\frac{BH}{BC} + \frac{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 $: AB.AC = AH.BC = 2S_{ABC}$)
