Đáp án:
Giải thích các bước giải:
a,Theo hệ thức lượng trong tam giác vuông
Ta có: $\frac{AB^4}{AC^4}$ =($\frac{AB^2}{AC^2}$) ²
=$\frac{(BH.BC)^2}{(CH.BC)^2}$ =$\frac{BH^2}{CH^2}$ =$\frac{BE.AB}{CF.AC}$ =$\frac{BE}{CF}$. $\frac{AB}{AC}$
⇒ $\frac{AB^3}{AC^3}$= $\frac{BE}{CF}$
b, Theo Py-ta-go ta có:
BC²=AB²+AC²
=BH²+AH²+CH²+AH²
=2AH²+BH²+CH²
=2AH²+BE²+EH²+HF²+CF²
=2AH²+BE²+CF²+EF²
=3AH²+BE²+CF² (Vì AH=EF)
c, BE$\sqrt[]{CH}$ +CF$\sqrt[]{BH}$
=$\frac{BH^2}{AB}$.$\sqrt[]{CH}$+$\frac{CH^2}{AC}$.$\sqrt[]{BH}$
=$\sqrt[]{BH.CH}$ ($\frac{\sqrt[]{BH^3}}{AB}$ +$\frac{\sqrt[]{CH^3}}{AC}$)
=$\sqrt[]{AH^2}$ ($\frac{\sqrt[]{BH^3.AC^2}+$\sqrt[]{CH^3.AB^2}}{AB.AC}$)
=AH.($\frac{\sqrt[]{BH^2.BH.BC.CH}+\sqrt[]{CH^2.CH.BC.BH}}{AH.BC}$ )
=AH.($\frac{\sqrt[]{BH^2.AH^2.BC}+\sqrt[]{CH^2.AH^2.BC}}{AH.BC}$ )
=AH.($\frac{BH.AH.\sqrt[]{BC}+CH.AH.\sqrt[]{BC}}{AH.BC}$
=AH.($\frac{BH+CH}{\sqrt[]{BC}}$ )
=AH.$\frac{BC}{\sqrt[]{BC}}$
=AH.$\sqrt[]{BC}$
