Question

Cine imi face aceste trei ex are 10 de la mine. :) ✅

Answer 1

 

$\displaystyle\bf\\1)~Comparati:numerele:\\\\a)\\\\9^{10}~si~27^7\\\\9^{10}=\Big(3^2\Big)^{10}=3^{2\times10}=3^{20}\\\\27^{7}=\Big(3^3\Big)^{7}=3^{3\times7}=3^{21}\\\\3^{20}<3^{21}\\\\\implies~\boxed{\bf9^{10}<27^7}\\\\b)\\\\5,4(3)~si~5,(43)\\\\5,433333...~si~5,434343...\\\\Comparam~la~cifra~miimilor.\\3<4\\\\5,433333...<5,434343...\\\\\implies~\boxed{\bf5,4(3)<5,(43)}$

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$\displaystyle\bf\\c)\\\\4\sqrt{3}~si~3\sqrt{5}\\\\Le~ridicam~la~patrat.\\\Big(4\sqrt{3}\Big)^2~si~\Big(3\sqrt{5}\Big)^2\\\\16\times3~si~9\times5\\\\48>45\\\\\implies~\boxed{\bf4\sqrt{3}>3\sqrt{5}}$

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$\displaystyle\bf\\2)\\\\Scrieti~in~ordine~crescatoare~numerele:\\\\-4\sqrt{6};~~~-2\sqrt{23};~~~-7\sqrt{2}\\\\Ridicam~la~patrat~numerele~fara~semn.\\\\-\Big(4\sqrt{6}\Big)^2=-16\times6=-96\\\\-\Big(2\sqrt{23}\Big)^2=-4\times23=-92\\\\-\Big(7\sqrt{2}\Big)^2=-49\times2=-98\\\\-98~<~-96~<-~92\\\\\implies~~\boxed{\bf -7\sqrt{2}~<~-4\sqrt{6}~<~-2\sqrt{23}}$

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$\displaystyle\bf\\3)~Determinati~o~fractie~ordinara~cuprinsa~intre~fractiile:\\\\\frac{7}{8}~si~\frac{8}{9}\\\\Pasul~1:~~Aducem~fractiile~la~acelasi~numitor.\\\\\frac{7\times9}{8\times9}~si~\frac{8\times8}{9\times8}\\\\\frac{63}{72}~si~\frac{64}{72}\\\\Pasul~2:~~Amplificam~fractiile~cu~2.\\\\\frac{126}{144}~si~\frac{128}{144}\\\\\implies~intre~fract~avem~fractia:\\\\\frac{127}{144} \\\\\frac{126}{144}<\frac{127}{144}<\frac{128}{144}\\\\\implies~~\boxed{\bf\frac{7}{8}<\frac{127}{144}<\frac{8}{9}}$