2 replies to this topic

[studentlovemath]

Cho tam giác ABC vuông ở A. $AH\perp BC, HE\perp AB, HF\perp AC$ . Chứng minh: 
$\sqrt[3]{BC^{2}}=\sqrt[3]{CF^{2}}+\sqrt[3]{BE^{2}}$

[studentlovemath]

Hình

[baotranthaithuy]

$\frac{FC}{AC}=\frac{HC}{BC}$

$\Rightarrow FC=\frac{AC.HC}{BC}= \frac{AC.AC^{2}}{BC.BC}=\frac{AC^{3}}{BC^{2}}$

Tương tự $EB=\frac{AB^{3}}{BC^{2}}$

$\Rightarrow \sqrt[3]{CF^{2}}+\sqrt[3]{BE^{2}}

$=\sqrt[3]{(\frac{AC^{3}}{BC^{2}})^{2}}+\sqrt[3]{(\frac{AB^{3}}{BC^{2}})^{2}}$

$=\frac{AC^{2}+AB^{2}}{\sqrt[3]{BC^{4}}} =\frac{BC^{2}}{\sqrt[3]{BC^{4}}}=\sqrt[3]{BC^{2}}$