anh em coi thế nào nha giúp với góp ý
a) $BC^2=AB^2+AC^2=AH^2+BH^2+AH^2+CH^2=2AH^2+BH^2+CH^2$
b) $BH^4=BE^2.AB^2=BE^2.BH.BC\Rightarrow BH^3=BE^2.BC\Rightarrow BE^2=\dfrac{BH^3}{BC}$
$AH^4=BH^2.CH^2=BE.AB.CF.AC=BE.CF.BC.AH$
$\Rightarrow AH^3=BE.CF.BC$
Dễ dàng thấy $AEHF$ là hình chữ nhật nên $EF=AH$
$\Rightarrow BE.CF.BC=EF\color{red}{^3}$
c) $\dfrac{CF}{BE}.\dfrac{AC}{AB}=\dfrac{CH^2}{BH^2}=\dfrac{CH^2.BC^2}{BH^2.BC^2}=\dfrac{AC^4}{AB^4}$
$\Rightarrow \dfrac{CF}{BE}=\dfrac{AC^3}{AB^3}$
d) $\triangle ABH\sim \triangle CHF$ (g.g) $\Rightarrow \dfrac{AH}{CF}=\dfrac{BH}{HF}$
$\Rightarrow CF.BH=AH.HF$
$\Rightarrow CF.BH=AH.AE$
$\Rightarrow CF.BH=AH(AB-BE)$
$\Rightarrow CF.BH=AH.AB-AH.BE$
$\Rightarrow BE.AH+CF.BH=AH.AB$
$\Rightarrow BE\sqrt{AH^2}+CF\sqrt{BH^2}=AH\sqrt{AB^2}$
$\Rightarrow BE\sqrt{BH.CH}+CF{BH^2}=AH\sqrt{BH.BC}$
$\Rightarrow BE\sqrt{CH}+CF\sqrt{BH}=AH\sqrt{BC}$
e) $BE=BH.\cos B=AB.\cos^2 B=BC.\cos^3 B\Rightarrow BE^2=BC^2.\cos^6 B$
$\Rightarrow \sqrt[3]{BE^2}=\sqrt[3]{BC^2.\cos^6 B}=\sqrt[3]{BC^2}.\cos^2 B$
cmtt: $CF=\sqrt[3]{BC^2}.\cos^2 C=\sqrt[3]{BC^2}.\sin^2 B$
$\Rightarrow \sqrt[3]{BE^2}+\sqrt[3]{CF^2}=\sqrt[3]{BC^2}(\sin ^2 B+\cos^2 B)=\sqrt[3]{BC^2}$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
