Matematică, întrebare adresată de Utilizator anonim, 9 ani în urmă

Cine ma poate ajuta la exercițiu 2 punctele b c d e ! Va rogg !

Anexe:

Răspunsuri la întrebare

Răspuns de nonim
1
b)[tex]( \sqrt{3} )^{5} :(- \sqrt{3^{3} })-3* \sqrt{0,(3)}*( \frac{1}{ \sqrt{3} } )^{-2} = \\ 3^{2} \sqrt{3} :(-3 \sqrt{3} )-3* \sqrt{ \frac{1}{3} } *( \sqrt{3} )^{2} } = \\ -3- 3*\sqrt{ \frac{9}{3} } =-3-3 \sqrt{3} = \\ -3(1+ \sqrt{3}) [/tex]

c)[tex][ \sqrt{6561} * \frac{-3,4(1)}{-27,(9)} -( \frac{4}{3} ) ^{-2} *0,(4)]:( \frac{6}{ \sqrt{7} } )^{2} * \frac{1}{10} = \\ [/tex][tex][81* \frac{307}{90}: \frac{279}{10} -( \frac{3}{4})^{2} * \frac{4}{10} ] * \frac{36}{7} * \frac{1}{10} = \\ (81* \frac{307}{9*279} - \frac{9}{16} * \frac{2}{5} )* \frac{18}{35} = \\ ( \frac{307}{31} - \frac{9}{40} )* \frac{18}{35} = \\ \frac{12280-279}{1240} * \frac{18}{35} = \\ \frac{12001}{620} * \frac{9}{35} = \\ \frac{12001*9}{620*35} = \\ \frac{108009}{21700} =4 \frac{21209}{21700} [/tex]

d) \frac{1}{ \sqrt{10^{4} } } - \sqrt{9,61} :(3,1)^{-1} + \sqrt{0,0576} *(0,2) ^{-2} =   [tex] \frac{1}{10 ^{2} } -(3,1) ^{1-(-1)} +0,24*( \frac{1}{5}) ^{-2} = \frac{1}{10^{2} } *(3,1) ^{2} +0,24* 5^{2} \\ \frac{1}{100} *9,61+0,24*25= \frac{9,61}{100} +6=6 \frac{9,61}{100} [/tex]

e)[-1,2(5)+0,1(4)]*[12-1,1(5)-1,8(4)]=
[tex](- \frac{113}{90} + \frac{13}{90} ) *(12- \frac{52}{45} - \frac{83}{45} )= \\ \frac{-113+13}{90} *(12 +\frac{-52-83}{45} )= -\frac{100}{90} *(12- \frac{135}{45} )= - \frac{10}{9} *(12-3)= \\ - \frac{10}{9}*9=-10 [/tex]
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