Lời giải:
a. $\sqrt{5}-\sqrt{7}< 0 = \sqrt{6}-\sqrt{6}< \sqrt{6}-\sqrt{4}=\sqrt{6}-2$
b. $(\sqrt{10}+\sqrt{11}-\sqrt{7})-(\sqrt{10}+\sqrt{13}-\sqrt{5})$
$=(\sqrt{11}-\sqrt{13})+(\sqrt{5}-\sqrt{7})< 0+0=0$
$\Rightarrow \sqrt{10}+\sqrt{11}-\sqrt{7}< \sqrt{10}+\sqrt{13}-\sqrt{5}$
c.
$3.1024^2=3.(2^{10})^2=3.2^{20}> 2.2^{20}=2^{21}$
a:
5<7
=>\(\sqrt{5}< \sqrt{7}\)
=>\(\sqrt{5}-\sqrt{7}< 0\)(1)
6>4
=>\(\sqrt{6}>\sqrt{4}\)
=>\(\sqrt{6}-2>0\left(2\right)\)
Từ (1),(2) suy ra \(\sqrt{5}-\sqrt{7}< \sqrt{6}-2\)
b: \(\sqrt{11}-\sqrt{7}-\sqrt{13}+\sqrt{5}\)
\(=\left(\sqrt{11}-\sqrt{13}\right)-\left(\sqrt{7}-\sqrt{5}\right)\)
\(=-\left(\sqrt{13}-\sqrt{11}\right)-\left(\sqrt{7}-\sqrt{5}\right)\)
\(\sqrt{13}-\sqrt{11}>0;\sqrt{7}-\sqrt{5}>0\)
Do đó: \(\left(\sqrt{13}-\sqrt{11}\right)+\left(\sqrt{7}-\sqrt{5}\right)>0\)
=>\(-\left(\sqrt{13}-\sqrt{11}\right)-\left(\sqrt{7}-\sqrt{5}\right)< 0\)
=>\(\sqrt{11}-\sqrt{7}-\sqrt{13}+\sqrt{5}< 0\)
=>\(\sqrt{11}-\sqrt{7}< \sqrt{13}-\sqrt{5}\)
=>\(\sqrt{10}+\sqrt{11}-\sqrt{7}< \sqrt{10}+\sqrt{13}-\sqrt{5}\)
c: \(3\cdot1024^2=3\cdot\left(2^{10}\right)^2=3\cdot2^{20}\)
mà \(3\cdot2^{20}>2\cdot2^{20}=2^{21}\)
nên \(3\cdot1024^2>2^{21}\)
