good teacher

Mik làm sai rồi, mik nhầm ở chỗ định lý pitago, sorry bạn nha

nice

Áp dụng định lý Pitago ta có:

BC = a² + \(\left(a\sqrt{3}\right)^2\)

Ta có: S_ABC= \(\dfrac{1}{2}\)\(\left\{AH.\left[a^2+\left(a\sqrt{3}\right)^2\right]\right\}\)

S_ABC cũng có thể bằng: \(\dfrac{a.a\sqrt{3}}{2}\)

=> PT: \(\dfrac{1}{2}\left\{AH.\left[a^2+\left(a\sqrt{3}\right)^2\right]\right\}=\dfrac{a^2\sqrt{3}}{2}\)

⇔ \(\left\{AH.\left[a^2+\left(a\sqrt{3}\right)^2\right]\right\}=a^2\sqrt{3}\)

⇔ AH = \(\dfrac{a^2\sqrt{3}}{\left[a^2+\left(a\sqrt{3}\right)^2\right]}\) = \(\dfrac{a}{a^2}=\dfrac{1}{a}\)

good

Câu a sai bạn ơi, phân số mà

a. \(\dfrac{27}{15}-\dfrac{18}{15}\) = \(\dfrac{27-18}{15}\) = \(\dfrac{9}{15}\) = \(\dfrac{3}{5}\)

b. \(\dfrac{1}{2}+\dfrac{1}{6}-\dfrac{2}{3}=\dfrac{3}{6}+\dfrac{1}{6}-\dfrac{4}{6}=\dfrac{3+1-4}{6}=\dfrac{0}{6}=0\)

:))