Bardzo proszę o rozwiązanie. daje naj.... Question from @Mata16

December 2018 1 10 Report

Bardzo proszę o rozwiązanie. daje naj

ewusia2206 Zad. 1
a) $\sqrt[5]{25 ^{75} } = \sqrt[5]{(5 ^{2}) ^{75} } = \\ \\ = \sqrt[5]{5 ^{2*75} } = \sqrt[5]{5 ^{150} } = \\ \\ = \sqrt[5]{5 ^{3*50} } = \sqrt[5]{(5 ^{3}) ^{50}} = \\ \\ = \sqrt[5]{125 ^{50}} = 125 ^{ \frac{50}{5} } = \\ \\ = 125 ^{ 10 } = 125 ^{ \frac{100}{10} } = \sqrt[10]{125 ^{100} }$

b) $\sqrt[4]{2} * \sqrt[6]{3} * \sqrt[12]{3} = 2 ^{ \frac{1}{4} } * 3 ^{ \frac{1}{6} } * 3 ^{ \frac{1}{12} } = \\ \\ = 2 ^{ \frac{1}{4} } * 3 ^{ \frac{1}{6} + \frac{1}{12} } = 2 ^{ \frac{1}{4} } * 3 ^{ \frac{2}{12} + \frac{1}{12} } = \\ \\ = 2 ^{ \frac{1}{4} } * 3 ^{ \frac{3}{12}} = 2 ^{ \frac{1}{4} } * 3 ^{ \frac{1}{4}} = \\ \\ =(2*3) ^{ \frac{1}{4} } = 6^{ \frac{1}{4} } = \sqrt[4]{6 ^{1} } = \sqrt[4]{6}$

Zad. 2
$243^{11} -81 ^{14} + 7x = 9 ^{27} \\ \\ ( 3^{5}) ^{11} - ( 3^{4} )^{14} + 7x = (3 ^{2} )^{27} \\ \\3 ^{5*11} - 3^{4*14} + 7x = 3^{2*27} \\ \\ 3^{55} - 3^{56} + 7x = 3^{54} \\ \\ 7x = 3^{54} -3^{55} + 3^{56} \\ \\ 7x = 3^{54} -3^{54+1} + 3^{54+2} \\ \\ 7x = 3^{54} -3^{54}*3 ^{1} + 3^{54} * 3^{2} \\ \\ 7x = 3^{54}(1 - 3 + 3 ^{2} ) \\ \\ 7x = 3^{54}(1-3+9) \\ \\ 7x = 3^{54}(-2+9) \\ \\ 7x=3^{54} * 7 \\ \\ x=3^{54}$

Zad. 3
$a = 3 ^{ \sqrt{2}-2 } , b = 3 ^{ \sqrt{2}} \\ \\ 9a =9 * 3 ^{ \sqrt{2}-2 } \\ \\ 9a = 3^{2} * 3 ^{ \sqrt{2}-2 } \\ \\ 9a =3 ^{2+ \sqrt{2}-2} \\ \\ 9a = 3 ^{\sqrt{2}} \\ \\ 9a = b$

Zad. 4
$a = \sqrt[3]{2 \frac{10}{27} } * (\frac{4}{3} ) ^{-2} = \sqrt[3]{ \frac{64}{27} } * ( \frac{3}{4} ) ^{2} = \frac{4}{3} * \frac{9}{16} = \frac{36}{48} = \frac{3}{4}$

$b) -0,5^{2} * (\frac{1}{3}) ^{-1} = -0,25 * (\frac{3}{1})^{1}= -0,25 * 3 = -0,75 = - \frac{3}{4}$

$c) \sqrt[3]{-( \frac{8}{3 \sqrt{3} }) ^{2} } = \sqrt[3]{- \frac{64}{3 ^{2}*( \sqrt{3}) ^{2} } } = \sqrt[3]{- \frac{64}{9*3} } = \sqrt[3]{- \frac{64}{27} } = - \frac{4}{3} = -1 \frac{1}{3}$

$d= \frac{0,3 ^{-1}*0,4 ^{2} }{2,5 ^{-1} } = \\ \\ = \frac{ (\frac{3}{10}) ^{-1}* (\frac{4}{10}) ^{2} }{(2 \frac{5}{10}) ^{-1} } = \\ \\ =\frac{ (\frac{10}{3})* (\frac{2}{5}) ^{2} }{(2 \frac{1}{2}) ^{-1} } = \\ \\ =\frac{\frac{10}{3}* \frac{4}{25} }{( \frac{5}{2}) ^{-1} }= \\ \\ =\frac{\frac{40}{75} }{( \frac{2}{5}) }= \\ \\ =\frac{40}{75} : \frac{2}{5} = \frac{40}{75} * \frac{5}{2} = \\ \\ =\frac{8}{15}* \frac{5}{2} = \frac{40}{30} = \frac{4}{3}=1 \frac{1}{3}$

W kolejności rosnącej: $- 1\frac{1}{3} ; - \frac{3}{4}; \frac{3}{4}; 1 \frac{1}{3}$ , czyli c<b<a<d

Zad. 5
$( \sqrt{2} - \sqrt{6} ) ^{2} +4 \sqrt{3} = \\ \\ =( \sqrt{2} )^{2} - 2 * \sqrt{2} * \sqrt{6} +( \sqrt{6} ) ^{2} + 4 \sqrt{3} = \\ \\ =4-2 \sqrt{12} +6+4 \sqrt{3} = \\ \\ =4-2 \sqrt{4*3} +6+4 \sqrt{3} = \\ \\ =4-2*2 \sqrt{3} +6+4 \sqrt{3} = \\ \\ =4-4 \sqrt{3} +6+4 \sqrt{3} = 10$

Liczba 10 jest liczbą naturalną.

Zad. 6
$2010^{2010} = (67*30) ^{2010} = \\ \\ =67 ^{2010} * 30^{2010} = 67^{201*10} * 30^{2010} = \\ \\ =67^{201}* 67^{10} * 30^{2010}$
Liczba $2010 ^{2010}$ zawiera czynnik $67 ^{201}$ , zatem jest przez niego podzielna.

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