Odpowiedź:
a) [tex]\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5[/tex]
b ) [tex]\sqrt{9\sqrt{16} } =\sqrt{9 \cdot 4}=\sqrt{36}=6[/tex]
c) dobrze
d) [tex]\sqrt{\frac{98}{128} } =\frac{\sqrt{98} }{\sqrt{128}}=\frac{\sqrt{2 \cdot 49} }{\sqrt{2^7} } =\frac{7\sqrt{2} }{\sqrt{2^2 \cdot 2^5} } =\frac{7\sqrt{2} }{2\sqrt{32} }=\frac{7\sqrt{2}}{2\sqrt{2^2 \cdot 2^3} } =\frac{7\sqrt{2} }{4\sqrt{8} }=\frac{7\sqrt{2} }{4\sqrt{2} \cdot 2 } =\frac{7}{8}[/tex]
