Odpowiedź:
[tex]zad.a~~~~?=8\\\\zad.b~~~~?=\dfrac{4}{3} \\\\zad.c~~~~?=\dfrac{3}{8} \\\\zad.d~~~~?=25[/tex]
Szczegółowe wyjaśnienie:
[tex]zad.a\\\\\dfrac{1}{2} \sqrt{x} =\sqrt{\dfrac{1}{4} } \cdot \sqrt{x} =\sqrt{ \dfrac{x}{4} } \\\\\sqrt{2} =\sqrt{ \dfrac{x}{4} } ~~\mid ()^{2} \\\\2=\dfrac{x}{4} ~~\mid \cdot~ 4\\\\x=8\\\\sprawdzam:\\\\\dfrac{1}{2} \sqrt{8} =\dfrac{1}{2} \sqrt{4\cdot 2}=\dfrac{1}{2} \cdot 2\sqrt{2}=\sqrt{2} ~~~~\Rightarrow~~L=P[/tex]
[tex]zad.b\\\\2\sqrt{3} =3\sqrt{x} \\\\\sqrt{2^{2} } \cdot \sqrt{3} =\sqrt{3^{2} } \cdot \sqrt{x} \\\\\sqrt{12} =\sqrt{9x} ~~\mid~()^{2} \\\\9x=12~~\mid \div ~9\\\\x=\dfrac{12}{9} \\\\x=\dfrac{4}{3} \\\\sprawdzam:\\\\3\sqrt{\dfrac{4}{3} } =\dfrac{6}{\sqrt{3} } \cdot \dfrac{\sqrt{3} }{\sqrt{3} } =\dfrac{6\sqrt{3} }3}=2\sqrt{3} ~~~~\Rightarrow~~L=P[/tex]
[tex]zad.c\\\\\dfrac{3}{2} \sqrt[3]{3} =3\sqrt[3]{x} ~~\mid \div~3\\\\\dfrac{1}{2} \sqrt[3]{3} =\sqrt[3]{x}\\\\\sqrt[3]{(\dfrac{1}{2} )^{3} } \cdot \sqrt[3]{3} =\sqrt[3]{x}\\\\\sqrt[3]{\dfrac{3}{8} } =\sqrt[3]{x} ~~\mid ()^{3} \\\\x=\dfrac{3}{8} \\\\sprawdzam:\\\\3\sqrt[3]{\dfrac{3}{8} } =3\sqrt[3]{\dfrac{3}{2^{3} } }=\dfrac{3}{2} \sqrt[3]{3} ~~~~\Rightarrow~~L=P[/tex]
[tex]zad.d\\\\\sqrt[3]{\dfrac{1}{5} } =\dfrac{1}{5} \sqrt[3]{x} \\\\\sqrt[3]{\dfrac{1}{5} } =\sqrt[3]{(\frac{1}{5} )^{3} } \cdot \sqrt[3]{x} \\\\\sqrt[3]{\dfrac{1}{5} } =\sqrt[3]{\dfrac{x}{125} } ~~\mid ()^{3} \\\\\dfrac{1}{5} =\dfrac{x}{125} ~~\mid \cdot 125\\\\x=25\\\\sprawdzam:\\\\\dfrac{1}{5} \sqrt[3]{25} =\sqrt[3]{(\frac{1}{5} )^{3} } \cdot \sqrt[3]{25}=\sqrt[3]{\dfrac{1}{125} \cdot 25}=\sqrt[3]{\dfrac{1}{5} } ~~~~\Rightarrow~~L=P[/tex]
