[tex]Zadanie\ 5\\\\a)\\\\\sqrt{3^{2}+4^{2} }=\sqrt{9+16}=\sqrt{25}=5\\\\b)\\\\ \sqrt{9\sqrt{16} }= \sqrt{9\cdot4}=\sqrt{36}=6\\\\c)\\\\\sqrt{16}+\sqrt{36}= 4+6=10\\\\d)\\\\ \sqrt{\frac{98}{128} }=\sqrt{\frac{49}{64} } =\frac{7}{8}[/tex]
