$x^3=\sqrt{4-x^2}+3x$
$GPT: x^3=\sqrt{4-x^2}+3x$
#1
Posted 24-01-2014 - 20:46
#2
Posted 24-01-2014 - 20:55
$x^3=\sqrt{4-x^2}+3x$
PT $< = > (x^3-3x)^2=4-x^2< = > x^6-6x^4+10x^2-4=0< = > t^3-6t^2+10t-4=0< = > t=2,t=2+\sqrt{2},t=2-\sqrt{2}$
- Yagami Raito and VNSTaipro like this
#3
Posted 24-01-2014 - 21:05
$x^3=\sqrt{4-x^2}+3x$
Điều kiện $-2\leq x \leq 2$
Đặt $x=2 \cos t t \in[0; \pi]$ ta có phương trình :
$8\cos^3 t -6 \cos t= 2 \sqrt{1-\cos^2 t}$
$\Leftrightarrow \cos 3t= \sin t$
$\Leftrightarrow t=\frac{\pi}{2}+k\pi; t=-\frac{\pi}{4}+k\frac{\pi}{4}$...
- Yagami Raito likes this
#4
Posted 28-01-2014 - 09:00
PT $< = > (x^3-3x)^2=4-x^2< = > x^6-6x^4+10x^2-4=0< = > t^3-6t^2+10t-4=0< = > t=2,t=2+\sqrt{2},t=2-\sqrt{2}$
Điều kiện $-2\leq x \leq 2$
Đặt $x=2 \cos t t \in[0; \pi]$ ta có phương trình :
$8\cos^3 t -6 \cos t= 2 \sqrt{1-\cos^2 t}$
$\Leftrightarrow \cos 3t= \sin t$
$\Leftrightarrow t=\frac{\pi}{2}+k\pi; t=-\frac{\pi}{4}+k\frac{\pi}{4}$...
thanks
- Hoang Tung 126 likes this
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