Question

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Answer 1

 

$\displaystyle\bf\\1a)\\\\a=3\sqrt{15}=\sqrt{9\times15}=\sqrt{135} \\b=7\sqrt{6}=\sqrt{49\times6}=\sqrt{294}\\135<294\\\implies~3\sqrt{15}<7\sqrt{6}~\implies~a<b\\\\1b)\\3\sqrt{8}=\sqrt{x}\\\sqrt{9\times8}=\sqrt{x}\\\\\sqrt{72}=\sqrt{x}\\x=72\\\\1c)\\\frac{\sqrt{128}}{4}=\frac{\sqrt{64\times2}}{4}=\frac{8\sqrt{2}}{4}=2\sqrt{2}$

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$\displaystyle\bf\\2a)\\Produsul~dintre~un~nr~rational~si~un~nr~irational=irational,\\cu~exceptia~cazului~in~care~nr~rational=0.\\\\\sqrt{3}(a-2)\in Q;~a\in~q\\\implies~(a-2)=0\\\implies~a=2\\\\ 2b)\\\frac{\sqrt{17^2-8^2}}{3}=\frac{\sqrt{289-64}}{3}=\frac{\sqrt{289-64}}{3}=\frac{\sqrt{225}}{3}=\frac{15}{3}=5\in Q~...~(A)\\\\2c)\\n<2\sqrt{5}<n+1~~~(Ridicam~la~patrat)\\n^2<(2\sqrt{5})^2<(n+1)^2\\\\n^2<20<(n+1)^2\\16<20<25\\4<2\sqrt{5}<5\\\implies~n=4\in N~~si~~n+1=4+1=5$