Giúp mk bài này với mọi người!

B

buichianh18896

$\begin{array}{l}
y' = \frac{8}{{{{(x + 2)}^2}}}\\
pttt\_dt(d)\_qua\_A(a;2)\_voi\_hsg\_k:y = k(x - a) + 2\\
(d)\_la\_tt\_cua\_dths\_khi:\\
k = \frac{8}{{{{(x + 2)}^2}}}\\
va`:\frac{8}{{{{(x + 2)}^2}}}(x - a) + 2 = \frac{{2x - 4}}{{x + 2}}\\
\Leftrightarrow 16x - 8a + 16 = 0\\
B({x_0};\frac{{2{x_0} - 4}}{{{x_0} + 2}}) \in (d) \Rightarrow 16{x_0} - 8a + 16 = 0 \Rightarrow a = 2{x_0} + 2\\
\Rightarrow AB = \sqrt {{{({x_o} + 2)}^2} + {{(\frac{8}{{{x_0} + 2}})}^2}} = 2\sqrt 5 \\
Dat:t = {({x_0} + 2)^2} = t,t \ge 0\\
\Rightarrow t + \frac{{64}}{t} = 20 \Leftrightarrow {t^2} - 20t + 64 = 0\\
giai\_pt...................
hoac:t = 4\\hoac:t = 16\end{array}$
 
Last edited by a moderator:
You must log in or register to reply here.
Top Bottom