Giải phương trình sau: $$ x=\sqrt{x-\displaystyle \frac{1}{x}}+\sqrt{1-\displaystyle \frac{1}{x}} $$
Giải phương trình: $ x=\sqrt{x-\displaystyle \frac{1}{x}}+\sqrt{1-\displaystyle \frac{1}{x}} $
Started By Alexman113, 12-04-2012 - 18:49
#1
Posted 12-04-2012 - 18:49
KK09XI~ Nothing fails like succcess ~
#2
Posted 12-04-2012 - 19:24
$$ x=\sqrt{x-\displaystyle \frac{1}{x}}+\sqrt{1-\displaystyle \frac{1}{x}} $$Giải phương trình sau:$$ x=\sqrt{x-\displaystyle \frac{1}{x}}+\sqrt{1-\displaystyle \frac{1}{x}} $$
$\Leftrightarrow x=\frac{\sqrt{{{x}^{2}}-1}+\sqrt{x-1}}{\sqrt{x}}$
$\Leftrightarrow x\sqrt{x}=\sqrt{x-1}(\sqrt{x+1}+1)=\sqrt{x-1}\frac{x}{\sqrt{x+1}-1}$
$\Leftrightarrow \sqrt{x}(\sqrt{x+1}-1)=\sqrt{x-1}$
$\Leftrightarrow \sqrt{{{x}^{2}}+x}=\sqrt{x}+\sqrt{x-1}$
$\Leftrightarrow {{x}^{2}}+x=2x-1+2\sqrt{{{x}^{2}}-x}4$
$\Leftrightarrow {{x}^{2}}-x-2\sqrt{{{x}^{2}}-x}+1=0$
$\Leftrightarrow \sqrt{{{x}^{2}}-x}=1$
$\Leftrightarrow \left[ \begin{align} x=\dfrac{1-\sqrt{5}}{2} \\ x=\dfrac{1+\sqrt{5}}{2} \end{align} \right.$
Thế vào phương trình ban đầu thấy nghiệm $x=\frac{1+\sqrt{5}}{2}$ thỏa mãn.
Edited by huynhmylinh, 12-04-2012 - 19:25.
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