Szczegółowe wyjaśnienie:
[tex]1.\\1)\ (\sqrt{36}-\sqrt{16})\cdot2=(*)\\\\\sqrt{36}=6\ bo\ 6^2=36\\\sqrt{16}=4\ bo\ 4^2=16\\\\(*)=(6-4)\cdot2=2\cdot2=4\\\boxed{\bold{P}}\\2)\ \sqrt{9^2}+\left(\sqrt6\right)^2=(*)\\\\Wzory:\ \sqrt{a^2}=a\ i\ \left(\sqrt{a}\right)^2=a\ dla\ a\geq0\\\\(*)=9+6=15\\\boxed{\bold{P}}[/tex]
[tex]2.\\ \sqrt{x}=1,2\to x=1,2^2\to x=1,44\\\boxed{\bold{C.\ 1,44}}[/tex]
[tex]3.\\\sqrt{1\dfrac{9}{16}}=\sqrt{\dfrac{25}{16}}=\dfrac{\sqrt{25}}{\sqrt{16}}=(*)\\\\\sqrt{25}=5\ bo\ 5^2=25\\\sqrt{16}=4\ bo\ 4^2=16\\\\(*)=\dfrac{5}{4}=1\dfrac{1}{4}\\\boxed{\bold{C.\ 1\dfrac{1}{4}}}[/tex]
[tex]4.\\\sqrt{\dfrac{9}{16}}=\dfrac{3}{4}\ bo\ \left(\dfrac{3}{4}\right)^2=\dfrac{9}{16}\\\boxed{\bold{D.\ \dfrac{3}{4}}}[/tex]
[tex]5.\\\sqrt{28}=\sqrt{4\cdot7}=\sqrt4\cdot\sqrt7=2\sqrt7\\\boxed{\bold{B.\ 2\sqrt7}}[/tex]
