Question

Ex 3 și 4 urgent va roogg!! DAU COROANA​

Answer 1

 

$\displaystyle\bf\\3)\\\\\sqrt{2}\cdot\sqrt{2^2}\cdot\sqrt{2^3}\cdot\sqrt{2^4}=\sqrt{2\cdot2^2\cdot2^3\cdot2^4}}=\sqrt{2^{1+2+3+4}}=\sqrt{2^{10}}\\\\\sqrt{2^{10}}=2^5=32\in N\subset Z\subset Q~~(este~numar~rational)\\\\\\4)\\\\\sqrt{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10}=\sqrt{10!}\\\\Descompunem~numarul~10!~(factorial)~in~factori~primi.\\Calculam~exponentii~factorilor~primi~mai~mici~decat~10.$

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$\displaystyle\bf\\2^n\\n=[10:2]+[10:2^2]+[10:2^3]=\\=[10:2]+[10:4]+[10:8]=5+2+1=8\\\implies~2^8\\\\3^n\\n=[10:3]+[10:3^2]=[10:3]+[10:9]=3+1=4\\\implies~~3^4\\\\5^n\\n=[10:5]=2\\\implies~~5^2\\\\7^n\\n=[10:7]=1\\\implies~~7^1$

Paranteza patrata inseamna "Partea intreaga a rezultatului impartirii"

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$\displaystyle\bf\\10!=1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10=2^8\cdot3^4\cdot5^2\cdot7\\\\\sqrt{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10}=\\\\=\sqrt{2^8\cdot3^4\cdot5^2\cdot7)}=2^4\cdot3^2\cdot 5\sqrt{7}= \\\\=16\cdot9\cdot5\sqrt{7}=720\sqrt{7} \in R-Q~~~(Numar~irational)$