Phương trình vô tỉ

Question

$\sqrt{2-2\sqrt{1-x^2}}=x(1+\sqrt{1-x^2})$

Nero · Answer 1 · 6/18/2014 11:22:09 AM

ĐK:

$-1\leq x\leq 1$

Đặt

$x=sint,t\in [-\frac{\pi}{2};\frac{\pi}{2}]\Rightarrow \sqrt{1-x^2}=\sqrt{1-sin^2t}=\left| {cost} \right|=cost$

Pt

$\Leftrightarrow \sqrt{2(1-cost)}=sint(1+cost)$

$\Leftrightarrow \sqrt{4sin^2\frac{t}{2}}=2sint.cos^2\frac{t}{2}$

$\Leftrightarrow \left| {sin\frac{t}{2}} \right|=2sin\frac{t}{2}.cos^3\frac{t}{2} (1)$

Với $t\in [-\frac{\pi}{2};0)\Rightarrow \frac{t}{2}\in [-\frac{\pi}{4};0)\Rightarrow \begin{cases}sin\frac{t}{2}<0 \\ cos\frac{t}{2}>0 \end{cases}$ thì $(1)\Leftrightarrow -sin\frac{t}{2}=2sin\frac{t}{2}cos^3\frac{t}{2}\Leftrightarrow sin\frac{t}{2}(1+2cos^3\frac{t}{2})=0 (VN)$
Với $t\in [0;\frac{\pi}{2}]\Rightarrow \frac{t}{2}\in [0;\frac{\pi}{4}]\Rightarrow \begin{cases}sin\frac{t}{2}\geq 0\\ cos\frac{t}{2}\leq 0\end{cases}$ thì $(1)\Leftrightarrow sin\frac{t}{2}=2sin\frac{t}{2}cos^3\frac{t}{2}\Leftrightarrow sin\frac{t}{2}(1-2cos^3\frac{t}{2})=0\Leftrightarrow sin\frac{t}{2}=0\vee cos\frac{t}{2}=\frac{1}{\sqrt[3]{2}}$
Với $\begin{cases}sin\frac{t}{2}=0 \\ t\in [0;\frac{\pi}{2}]\end{cases}\Rightarrow \begin{cases}t=k2\pi \\ t\in[0;\frac{\pi}{2}] \end{cases}\Leftrightarrow t=0\Leftrightarrow x=sint=0$
Với $\begin{cases}cos\frac{t}{2}=\frac{1}{\sqrt[3]{2}} \\ t\in [0;\frac{\pi}{2}]\end{cases}\Rightarrow \begin{cases}sin\frac{t}{2}=\sqrt{1-\frac{1}{\sqrt[3]{4}}}=\frac{\sqrt{\sqrt[3]{4}-1}}{\sqrt[3]{2}} \\ t\in [0;\frac{\pi}{2}]\end{cases}\Leftrightarrow x=sint=2sintcost=\sqrt{\sqrt[3]{4}-1}$