Giải phương trình : 
$$x^3-3x=\sqrt{x+2}$$
nghiệm lẻ :3 khó chịu thế –  tran85295 02-08-16 11:57 PM
đây ko phải thú cưng nhá =.= –  Confusion 02-08-16 11:16 PM
vầy bình phương :)) –  Confusion 02-08-16 11:15 PM
đúng r đó cưng ko tin giải thử đi –  ★·.·´¯`·.·★Poseidon★·.·´¯`·.·★ 02-08-16 10:12 PM
thế ak ==" –  Confusion 02-08-16 10:11 PM
liên hợp dk mỗi nghiệm con này 3 nghiệm lận –  ★·.·´¯`·.·★Poseidon★·.·´¯`·.·★ 02-08-16 10:11 PM
hehe cứ thử đi bà –  ★·.·´¯`·.·★Poseidon★·.·´¯`·.·★ 02-08-16 10:10 PM
liên hợp phát ra luk đấy thây ==" –  Confusion 02-08-16 10:06 PM
$pt\Leftrightarrow (\sqrt{x+2}-2)(x+1+\sqrt{x+2})(x+1+x\sqrt{x+2})=0$
Hai ngoặc đầu giải dễ dàng và cho 2 nghiệm $x_1=2,x_2=\frac{-1-\sqrt5}{2}$
$x+1+x\sqrt{x+2}=0\Leftrightarrow \sqrt{x+2}=\frac{-(x+1)}x$
$\Leftrightarrow \begin{cases}-1 \le x<0\\ x^3 +x^2=2x+1 =0 \;(1)\end{cases} $
$(1)\Leftrightarrow \left(x+\frac 13 \right)^3-\frac 73\left(x+\frac 13 \right)-\frac 7{27}=0$
$x+\frac 13 \longrightarrow \frac{2\sqrt 7 \cos \alpha }{3},\alpha \in [0,\pi]$
$\Rightarrow \frac{56\sqrt 7}{27} \cos^3 \alpha-\frac{14\sqrt 7}{9} \cos \alpha=\frac 7{27}$
$\Leftrightarrow 4\cos^3a -3\cos \alpha=\frac{\sqrt 7}{14}$
$\Leftrightarrow \cos 3\alpha=\frac{\sqrt 7}{14}$
$\Leftrightarrow 3 \alpha=k2\pi + \arccos \frac{\sqrt 7}{14}  (2)$ hoặc $3 \alpha =k2\pi -\arccos \frac{\sqrt 7}{14} (3)$
Vì $\alpha \in [0,\pi]$ nên từ $(2)$ chọn $k=0,k=1$ (loại vì đk $-1 \le x <0$)
Từ $(3)$ ta chọn $k=1$ (nhận)
Và ta có $x_3= \frac{2\sqrt 7 \cos \left(\dfrac{2\pi-\arccos \frac{\sqrt 7}{14}}{3} \right)-1}{3}$




Oh my god!! –  ★·.·´¯`·.·★Poseidon★·.·´¯`·.·★ 03-08-16 07:24 PM
ĐK: $x\geq -2$
$\bigstar$ Với $x>2\rightarrow VT=x+x(x^2-4)>2x>\sqrt{x+2}\rightarrow $ PT vô nghiệm.
$\bigstar$ Với $x \in [-2;2],$ set $x=2cost$ $(t\in[0;\pi])$
$\rightarrow \sqrt{x+2}=\sqrt{2(1+cost)}=\sqrt{2.2cos^2\frac{t}{2}}=2|cos\frac{t}{2}|=2cos\frac{t}{2}$ ( do $t \in [0;\frac{\pi}{4}])$
PT $\Leftrightarrow 2(4cos^3t-3cost)=2cos\frac{t}{2}$
$\Leftrightarrow 8cos^3t-6cost=2cos\frac{t}{2}$
$\Leftrightarrow cos3t=cos\frac{t}{2}$
$\Leftrightarrow 3t=\frac{t}{2}+k2\pi$ v $3t=\frac{-t}{2}+k2\pi$ $(k \in Z)$
$\Leftrightarrow t=\frac{k4\pi}{5}$ v $t=\frac{k4\pi}{7}$
Do $\left\{ \begin{array}{l} t \in [0;\pi]\\ k \in Z \end{array} \right.\Rightarrow t \in {0;\frac{4 \pi}{7};\frac{4 \pi }{5}}$
$\Rightarrow x={2;2cos\frac{4 \pi }{7};2cos\frac{4 \pi }{5}}$
ver lơ là cái j? –  Confusion 03-08-16 07:50 PM
toàn các thánh trâu ver lơ –  ★·.·´¯`·.·★Poseidon★·.·´¯`·.·★ 03-08-16 07:24 PM

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