1)
[tex]\frac{1}{3}\sqrt{27} =\frac{1}{3}\sqrt{3*9}=\frac{1}{3}*3\sqrt{3} =\sqrt{3}[/tex]
[tex]\sqrt{2a^7}=a^3\sqrt{2a}[/tex]
2)
[tex]5\sqrt{a}=\sqrt{25a} \\3k\sqrt{\frac{1}{k^2} } =k\sqrt{\frac{9}{k^2} }[/tex]
3)
[tex]\frac{4}{\sqrt{xy} } =\frac{4*\sqrt{xy} }{\sqrt{xy}*\sqrt{xy} } =\frac{4\sqrt{xy} }{xy}[/tex]
[tex]\frac{3}{\sqrt{5} +\sqrt{2} }=\frac{3(\sqrt{5}-\sqrt{2} )}{(\sqrt{5} +\sqrt{2})(\sqrt{5} -\sqrt{2} ) } = \frac{3(\sqrt{5} -\sqrt{2} )}{5-2} =\sqrt{5} -\sqrt{2}[/tex]
4)
[tex]y=\sqrt{x} \\y=\sqrt{9}\\ y=3 \\[/tex]
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Answers & Comments
1)
[tex]\frac{1}{3}\sqrt{27} =\frac{1}{3}\sqrt{3*9}=\frac{1}{3}*3\sqrt{3} =\sqrt{3}[/tex]
[tex]\sqrt{2a^7}=a^3\sqrt{2a}[/tex]
2)
[tex]5\sqrt{a}=\sqrt{25a} \\3k\sqrt{\frac{1}{k^2} } =k\sqrt{\frac{9}{k^2} }[/tex]
3)
[tex]\frac{4}{\sqrt{xy} } =\frac{4*\sqrt{xy} }{\sqrt{xy}*\sqrt{xy} } =\frac{4\sqrt{xy} }{xy}[/tex]
[tex]\frac{3}{\sqrt{5} +\sqrt{2} }=\frac{3(\sqrt{5}-\sqrt{2} )}{(\sqrt{5} +\sqrt{2})(\sqrt{5} -\sqrt{2} ) } = \frac{3(\sqrt{5} -\sqrt{2} )}{5-2} =\sqrt{5} -\sqrt{2}[/tex]
4)
[tex]y=\sqrt{x} \\y=\sqrt{9}\\ y=3 \\[/tex]