So sánh các số: a) √7 – √2 và 1 b) √8 – √5 và √7 + √6 19/09/2021 Bởi Sadie So sánh các số: a) √7 – √2 và 1 b) √8 – √5 và √7 + √6
a, $(\sqrt{7}-\sqrt{2})^2= 7-2\sqrt{14}+2= 9-2\sqrt{14}$ $\Rightarrow (\sqrt7-\sqrt2)^2-1= 8-2\sqrt{14}$ Ta có $8^2=64; (2\sqrt{14})^2=56$ $\Rightarrow 8^2> (2\sqrt{14})^2$ $\Leftrightarrow 8>2\sqrt{14}$ $\Leftrightarrow 8-2\sqrt{14}>0$ $\Leftrightarrow (\sqrt7-\sqrt2)^2-1>0$ $\Leftrightarrow (\sqrt7-\sqrt2)^2>1$ $\Leftrightarrow \sqrt7-\sqrt2 > 1$ ($\sqrt7-\sqrt2>0\Rightarrow |\sqrt7-\sqrt2|=\sqrt7-\sqrt2$) b, $(\sqrt8-\sqrt5)^2= 8+5-2\sqrt{8.5}=13-2\sqrt{40}$ $(\sqrt7+\sqrt6)^2=7+6+2\sqrt{7.6}=13+2\sqrt{42}$ $-2\sqrt{40} < 2\sqrt{42}$ $\Leftrightarrow 13-2\sqrt{40}<13+2\sqrt{42}$ $\Leftrightarrow (\sqrt8-\sqrt5)^2<(\sqrt7+\sqrt6)^2$ $\Leftrightarrow |\sqrt8-\sqrt5|<|\sqrt7+\sqrt6|$ $\Leftrightarrow \sqrt8-\sqrt5 < \sqrt7+\sqrt6$ Bình luận
a,
$(\sqrt{7}-\sqrt{2})^2= 7-2\sqrt{14}+2= 9-2\sqrt{14}$
$\Rightarrow (\sqrt7-\sqrt2)^2-1= 8-2\sqrt{14}$
Ta có $8^2=64; (2\sqrt{14})^2=56$
$\Rightarrow 8^2> (2\sqrt{14})^2$
$\Leftrightarrow 8>2\sqrt{14}$
$\Leftrightarrow 8-2\sqrt{14}>0$
$\Leftrightarrow (\sqrt7-\sqrt2)^2-1>0$
$\Leftrightarrow (\sqrt7-\sqrt2)^2>1$
$\Leftrightarrow \sqrt7-\sqrt2 > 1$ ($\sqrt7-\sqrt2>0\Rightarrow |\sqrt7-\sqrt2|=\sqrt7-\sqrt2$)
b,
$(\sqrt8-\sqrt5)^2= 8+5-2\sqrt{8.5}=13-2\sqrt{40}$
$(\sqrt7+\sqrt6)^2=7+6+2\sqrt{7.6}=13+2\sqrt{42}$
$-2\sqrt{40} < 2\sqrt{42}$
$\Leftrightarrow 13-2\sqrt{40}<13+2\sqrt{42}$
$\Leftrightarrow (\sqrt8-\sqrt5)^2<(\sqrt7+\sqrt6)^2$
$\Leftrightarrow |\sqrt8-\sqrt5|<|\sqrt7+\sqrt6|$
$\Leftrightarrow \sqrt8-\sqrt5 < \sqrt7+\sqrt6$