4 replies to this topic

[VNSTaipro]

Giải a)$(\sqrt{1+x}-1)(\sqrt{1-x}+1)=2x$
b)$(\sqrt{1-x}-1)(\sqrt{1+x}+1)=2x$

Edited by VNSTaipro, 28-02-2013 - 08:57.

[hoangtrong2305]

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

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

TXĐ: $x \in[-1;0]$

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

$\Leftrightarrow \frac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)(\sqrt{1-x}+1)}{(\sqrt{1+x}+1)}-2x=0$

$\Leftrightarrow \frac{x(\sqrt{1-x}+1)}{\sqrt{1+x}+1}-2x=0$

$\Leftrightarrow x[\frac{(\sqrt{1-x}+1)}{\sqrt{1+x}+1}-2]=0$

$\Leftrightarrow \begin{bmatrix} x=0\\ \frac{(\sqrt{1-x}+1)}{\sqrt{1+x}+1}-2=0 \end{bmatrix}$

Xét $\frac{(\sqrt{1-x}+1)}{\sqrt{1+x}+1}-2=0$

$\Leftrightarrow \sqrt{1-x}=2\sqrt{1+x}+1$

$\Leftrightarrow 1-x=4+4x+4\sqrt{1+x}+1$

$\Leftrightarrow 4\sqrt{1+x}=-(5x+4);x \in[-1;-\frac{4}{5}]$

$\Leftrightarrow 25x^{2}+36x+12=0$

$\Leftrightarrow ...........................$

[VNSTaipro]

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

TXĐ: $x \in[-1;0]$

$(\sqrt{1+x}-1)(\sqrt{1-x}+1)=2x$
$\Leftrightarrow \frac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)(\sqrt{1-x}+1)}{(\sqrt{1+x}+1)}-2x=0$
$\Leftrightarrow \frac{x(\sqrt{1-x}+1)}{\sqrt{1+x}+1}-2x=0$
$\Leftrightarrow x[\frac{(\sqrt{1-x}+1)}{\sqrt{1+x}+1}-2]=0$
$\Leftrightarrow \begin{bmatrix} x=0\\ \frac{(\sqrt{1-x}+1)}{\sqrt{1+x}+1}-2=0 \end{bmatrix}$
....

Anh có cách khác không, Dùng lượng giác á

[thukilop]

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

TXD: $x \in [-1;1]$

- Đặt $a=\sqrt{1+x}, b=\sqrt{1-x}$

Suy ra: $\left\{\begin{matrix}
a^{2}-b^{2}=2x & \\
a^{2}+b^{2}=2 &
\end{matrix}\right.$

-pt ban đầu tương đương: $(a-1)(b+1)=a^{2}-b^{2}$

-Giải hệ: $\left\{\begin{matrix}
a^{2}-b^{2}=(a-1)(b+1) & \\
a^{2}+b^{2}=2 &
\end{matrix}\right.$

<=> $\left\{\begin{matrix}
(a-b)(a+b-1)+1-ab=0 & \\
(a-b)^{2}=2(1-ab) &
\end{matrix}\right.$

Suy ra:$(a-b)(a+b-1)+\frac{(a-b)^{2}}{2}=0$ <=> $(a-b)(3a+b-2)=0$
<=>.........................................

KL: pt đã cho có nghiệm $\boxed{x=0,x=\frac{-24}{25}}$

p/s: câu b) chắc cũng tương tự thế

Edited by thukilop, 07-03-2013 - 02:44.

[nthoangcute]

Giải a)$(\sqrt{1+x}-1)(\sqrt{1-x}+1)=2x$
b)$(\sqrt{1-x}-1)(\sqrt{1+x}+1)=2x$

Làm như này cho ngắn:
a) $$(\sqrt{1+x}-1)(\sqrt{1-x}+1)-2x=\left( \sqrt {1+x}+2\,\sqrt {1-x}-3 \right) \left( \sqrt {1-x}+1
\right)$$
b) $$(\sqrt{1-x}-1)(\sqrt{1+x}+1)-2x= \left( \sqrt {1-x}-1 \right) \left( \sqrt {1+x}+2\,\sqrt {1-x}+3
\right) $$

Edited by nthoangcute, 17-03-2013 - 20:18.