So sánh các căn sau :
[tex]\sqrt{2006}-\sqrt{2005} và \sqrt{2005} -\sqrt{2004}[/tex]
[tex]\sqrt{1998}+\sqrt{2000} và 2\sqrt{1999}[/tex]
$\sqrt{2006}-\sqrt{2005}=(\sqrt{2006}-\sqrt{2005})(\sqrt{2006}+\sqrt{2005}).\dfrac{1}{\sqrt{2006}+\sqrt{2005}}=\dfrac{1}{\sqrt{2006}+\sqrt{2005}}$
$\sqrt{2005}-\sqrt{2004}=(\sqrt{2005}-\sqrt{2004})(\sqrt{2005}+\sqrt{2004}).\dfrac{1}{\sqrt{2005}+\sqrt{2004}}=\dfrac{1}{\sqrt{2005}+\sqrt{2004}}$
$\sqrt{2006}+\sqrt{2005}>\sqrt{2005}+\sqrt{2004}\Rightarrow \dfrac{1}{\sqrt{2006}+\sqrt{2005}}<\dfrac{1}{\sqrt{2005}+\sqrt{2004}}\Rightarrow \sqrt{2006}-\sqrt{2005}<\sqrt{2005}-\sqrt{2004}$
Xét hiệu $2\sqrt{1999}-(\sqrt{1998}+\sqrt{2000})=(\sqrt{1999}-\sqrt{1998})-(\sqrt{2000}-\sqrt{1999})$
$\sqrt{1999}-\sqrt{1998}=(\sqrt{1999}-\sqrt{1998})(\sqrt{1999}+\sqrt{1998}).\dfrac{1}{\sqrt{1999}+\sqrt{1998}}=\dfrac{1}{\sqrt{1999}+\sqrt{1998}}$
$\sqrt{2000}-\sqrt{1999}=(\sqrt{2000}-\sqrt{1999})(\sqrt{2000}+\sqrt{1999}).\dfrac{1}{\sqrt{2000}+\sqrt{1999}}=\dfrac{1}{\sqrt{2000}+\sqrt{1999}}$
$\sqrt{2000}+\sqrt{1999}>\sqrt{1999}+\sqrt{1998}\Rightarrow \dfrac{1}{\sqrt{2000}+\sqrt{1999}}<\dfrac{1}{\sqrt{1999}+\sqrt{1998}}\Rightarrow \sqrt{2000}-\sqrt{1999}<\sqrt{1999}-\sqrt{1998}$
$\Rightarrow 2\sqrt{1999}-(\sqrt{1998}+\sqrt{2000}) >0 \Rightarrow 2\sqrt{1999}>\sqrt{1998}+\sqrt{2000}$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
