Chủ đề này có 2 trả lời

[killerdark68]

 giải  $\sqrt{2x+\frac{2013x-1}{\sqrt{2-x^2}}}-\sqrt[3]{2014-\frac{2013x-1}{\sqrt{2-x^2}}}=\sqrt{x+2013}-\sqrt[3]{x+1}$

Bài viết đã được chỉnh sửa nội dung bởi Mikhail Leptchinski: 16-11-2014 - 22:39

[baotranthaithuy]

đặt $y=\frac{2013x-1}{\sqrt{2-x^{2}}}$

$\Rightarrow \sqrt{2x+y}-\sqrt[3]{2014-y}=\sqrt{x+2013}-\sqrt[3]{x+1} $

$*x+y>2013 \Rightarrow \left\{\begin{matrix} 2x+y>2013+x & \\ x+1>2014-y & \end{matrix}\right. $
$\Rightarrow VT>VP $

$*x+y<2013 \Rightarrow \left\{\begin{matrix} 2x+y<2013+x & \\ x+1<2014-y & \end{matrix}\right.$

$\Rightarrow VP>VT$
$ \Rightarrow x+y=2013$
$ \Leftrightarrow x+\frac{2013x-1}{\sqrt{2-x^{2}}}-2013=0$

$\Leftrightarrow x\sqrt{2-x^{2}}+2013(x-\sqrt{2-x^{2}})-1=0 t=x-\sqrt{2-x^{2}} $
$\Rightarrow x\sqrt{2-x^{2}}=\frac{2-t^{2}}{2} \Rightarrow \frac{2-t^{2}}{2}+2013t-1=0 $
$\Rightarrow \begin{bmatrix} t=0 & \\ t=4026 & \end{bmatrix} $
$@ t=0\Rightarrow x=\sqrt{2-x^{2}} \Leftrightarrow \left\{\begin{matrix} x\geq 0 & \\ x^{2}=2-x^{2} & \end{matrix}\right. .$

$\Leftrightarrow x=1 $
$@ *t=4026 \Rightarrow x-\sqrt{2-x^{2}}=4026 \Leftrightarrow x=\sqrt{2-x^{2}}+4026$

(vô nghiệm do $-\sqrt{2}\leq x\leq \sqrt{2} \Rightarrow VP>VT$)

[killerdark68]

đặt $y=\frac{2013x-1}{\sqrt{2-x^{2}}}$

$\Rightarrow \sqrt{2x+y}-\sqrt[3]{2014-y}=\sqrt{x+2013}-\sqrt[3]{x+1} $

$*x+y>2013 \Rightarrow \left\{\begin{matrix} 2x+y>2013+x & \\ x+1>2014-y & \end{matrix}\right. $
$\Rightarrow VT>VP $

$*x+y<2013 \Rightarrow \left\{\begin{matrix} 2x+y<2013+x & \\ x+1<2014-y & \end{matrix}\right.$

$\Rightarrow VP>VT$
$ \Rightarrow x+y=2013$
$ \Leftrightarrow x+\frac{2013x-1}{\sqrt{2-x^{2}}}-2013=0$

$\Leftrightarrow x\sqrt{2-x^{2}}+2013(x-\sqrt{2-x^{2}})-1=0 t=x-\sqrt{2-x^{2}} $
$\Rightarrow x\sqrt{2-x^{2}}=\frac{2-t^{2}}{2} \Rightarrow \frac{2-t^{2}}{2}+2013t-1=0 $
$\Rightarrow \begin{bmatrix} t=0 & \\ t=4026 & \end{bmatrix} $
$@ t=0\Rightarrow x=\sqrt{2-x^{2}} \Leftrightarrow \left\{\begin{matrix} x\geq 0 & \\ x^{2}=2-x^{2} & \end{matrix}\right. .$

$\Leftrightarrow x=1 $
$@ *t=4026 \Rightarrow x-\sqrt{2-x^{2}}=4026 \Leftrightarrow x=\sqrt{2-x^{2}}+4026$

(vô nghiệm do $-\sqrt{2}\leq x\leq \sqrt{2} \Rightarrow VP>VT$)

hoặc là đặt 4 ẩn cũng dc