Proszę to DOBRZE ZROBIĆ!!! Daje 264 pkt a zadań jest tylko 8!!!.... Question from @Luis405

December 2018 1 44 Report

Proszę to DOBRZE ZROBIĆ!!! Daje 264 pkt a zadań jest tylko 8!!!

Martius123 Zad.1
$\sqrt[3]{2 \frac{1}{4} } * \sqrt[3]{-1 \frac{1}{2} } - \sqrt[3]{1 \frac{7}{9} } * \sqrt[3]{-1 \frac{1}{3} } = \sqrt[3]{ \frac{9}{4} *(- \frac{3}{2}) } - \sqrt[3]{ \frac{16}{9} * (-\frac{4}{3}) } = \\ \sqrt[3]{- \frac{27}{8} } - \sqrt[3]{- \frac{64}{27} } =- \frac{3}{2} -(- \frac{4}{3})= \frac{4}{3} - \frac{3}{2}= \frac{8}{6} - \frac{9}{6} =- \frac{1}{6}$
$\frac{4 \sqrt{3} + \sqrt{27}}{ \sqrt{12}} = \frac{4 \sqrt{3} + 3\sqrt{3}}{2 \sqrt{3}} = \frac{4+3}{2} = \frac{7}{2} =3,5$
$2 \sqrt{75} -2 \sqrt{3} + \sqrt{300} = 2*5 \sqrt{3}- 2 \sqrt{3} +10 \sqrt{3} =18 \sqrt{3}$
$\frac{ 5^{8} : 5^{2} - 125^{2} }{ 25^{-3} : 5^{-2} } = \frac{ 5^{6}-5^{4}}{ (5^{2})^{-3}: 5^{-2} } = \frac{24* 5^{4} }{5^{-6-(-2)} }= \frac{24* 5^{4} }{ 5^{-4}} =24*5^{4-(-4)} =24* 5^{8}$
$\frac{ 3^{32}- 3^{30} }{ 3^{28} } = \frac{ 3^{30}}{ 3^{28} } *( 3^{2} -1)= 3^{2} *8=9*8=72$
$\frac{ 3^{5} + 27^{2} }{9* 3^{5} } = \frac{ 3^{5} +( 3^{3})^{2} }{ 3^{2}* 3^{5} } = \frac{3^{5} + 3^{6} }{ 3^{7}}= \frac{1+3}{ 3^{2} } = \frac{4}{9}$

Zad.2
$(4 \sqrt{5} + \sqrt{3})*(4 \sqrt{5} - \sqrt{3} )=(4 \sqrt{5})^{2}-( \sqrt{3} )^{2}=80-3=77$
$( \sqrt{6} -3 \sqrt{3})^{2}= ( \sqrt{6} )^{2} -2* \sqrt{6}*3 \sqrt{3} + (3\sqrt{3}) ^{2}=6-6 \sqrt{18} +27= \\ 33-18 \sqrt{2}$

Zad.3
Szukany obwód L
$c^{2}= x^{2} + y^{2} \\ c^{2} =(2 \sqrt{3} - \sqrt{2} )^{2}+(2 \sqrt{2} + \sqrt{3} )^{2} \\ c^{2}= (2 \sqrt{3}) ^{2} -2*2 \sqrt{3} * \sqrt{2} + (\sqrt{2} )^{2} + (2 \sqrt{2} )^{2} +2* 2\sqrt{2} * \sqrt{3} + ( \sqrt{3}) ^{2} \\ c^{2} =12-4 \sqrt{6} +2+8+4 \sqrt{6}+3 \\ c^{2} = 25 \\ c=5$

$L=x+y+c = 2 \sqrt{3} - \sqrt{2}+2 \sqrt{2} + \sqrt{3} +5= 5+ \sqrt{2}+ 3 \sqrt{3}$

Zad.4
$\frac{3}{ \sqrt{3} -1} = \frac{3( \sqrt{3} +1)}{( \sqrt{3} -1)( \sqrt{3} +1)} = \frac{3( \sqrt{3} +1)}{3-1} = \frac{3}{2} ( \sqrt{3} +1)$
$\frac{2 \sqrt{6} }{ \sqrt{3} -2} = \frac{2 \sqrt{6}( \sqrt{3}+2 ) }{(\sqrt{3} -2)(\sqrt{3} +2)} = \frac{2 \sqrt{6}( \sqrt{3}+2 )}{3-4} =-2 \sqrt{6}( \sqrt{3}+2 )$
$\frac{4}{5 \sqrt{3} } = \frac{4 \sqrt{3} }{5 \sqrt{3} \sqrt{3} } = \frac{4}{15} \sqrt{3}$
$\frac{-2}{3 \sqrt{2} } = \frac{-2 \sqrt{2} }{3 \sqrt{2} \sqrt{2} }=- \frac{2 \sqrt{2} }{6} =- \frac{ \sqrt{2} }{3}$

Zad.5
$3^{ \frac{8}{3} } * \sqrt[3]{ 9^{2} } =3^{ \frac{8}{3} }*[(3^{2})^{2}]^{ \frac{1}{3} }= 3^{ \frac{8}{3} }*3^{ \frac{4}{3} }=3^{ \frac{12}{3} }=3^{4}=81$
$( \frac{3+ \sqrt{3} }{ \sqrt{3}} )^{2}= \frac{(3+ \sqrt{3} )^{2}}{ (\sqrt{3} )^{2}} = \frac{9+6 \sqrt{3} +3}{3} = \frac{12+6 \sqrt{3} }{3} =4+2 \sqrt{3}$
$\sqrt{5+2 \sqrt{6} } * \sqrt{5-2 \sqrt{6} } = \sqrt{(5+2 \sqrt{6})(5-2 \sqrt{6})} = \sqrt{ 5^{2} -(2 \sqrt{6} )^{2}} = \\ \sqrt{25-24} = \sqrt{1}=1$

Zad.6
$(x+2y)-(x+2y)(2y-x)-y(3x-2y)= \\ x +2y +(x+2y)(x-2y)-3xy+2 y^{2} \\ x+2y+ x^{2} -4 y^{2} -3xy+2 y^{2} \\ x+2y+ x^{2} -3xy+2 y^{2} \\ x =\sqrt{3} \\ y=- \frac{1}{3} \\ wyrazenie=\sqrt{3} +2*(- \frac{1}{3} )+ (\sqrt{3}) ^{2} -3* \sqrt{3} *(- \frac{1}{3} )+2*(- \frac{1}{3} )^{2}= \\ \sqrt{3}- \frac{2}{3} +3+ \sqrt{3} - \frac{2}{9}=2 \sqrt{3} +2 \frac{1}{9}$

Zad.7
$||3 \sqrt{5} -2|- \sqrt{5}|=| 3\sqrt{5} -2- \sqrt{5}|=|2 \sqrt{5}-2 | =2 \sqrt{5} -2 \\ 
| \frac{3- \sqrt{2} }{2 \sqrt{2}-3 } |=| \frac{(3- \sqrt{2})*(2 \sqrt{2}+3 ) )}{(2 \sqrt{2}-3)*(2 \sqrt{2}+3)}|= \\ =| \frac{6 \sqrt{2}+9-4-3 \sqrt{2} }{8-9} |=|-(5+3 \sqrt{2})|=5+3 \sqrt{2}$
$| \sqrt{2}- \sqrt{5} |*|3 \sqrt{5}-10 |=-(\sqrt{2}- \sqrt{5}) *[-(3 \sqrt{5}-10)] \\ 3 \sqrt{10}-10 \sqrt{2}-15+10 \sqrt{5}$

Zad.8
$-0,2 ^{2} *( \frac{1}{2} )^{-1}=-0,04*2=0,08 \\ 
 \sqrt[3]{-( \frac{8}{3 \sqrt{3} })^{2}}= \sqrt[3]{-( \frac{64}{27} )} =- \frac{4}{3} \\ 
 \frac{(0,3)^-1*(0,4) ^{2} }{2,5^{-1}} = \frac{ \frac{10}{3} * \frac{16}{100} }{ \frac{2}{5} } = \frac{10}{3} * \frac{16}{100} * \frac{5}{2} = \frac{4}{3}$

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