Matematică, întrebare adresată de teodora75, 9 ani în urmă

I. 4 si II 1. va rog coroanaa

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teodora75: a 7 a

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Răspuns de Utilizator anonim
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\displaystyle  4).\frac{x}{ \sqrt{0,07(1)} }= \frac{ \sqrt{1,13(7)} }{ \sqrt{0,4096} } \Rightarrow  \frac{x}{  \sqrt{ \frac{71-7}{900} }  }  = \frac{\sqrt{ \frac{1137-113}{900}} }{ \sqrt{ \frac{4096}{10000} } }   \Rightarrow  \\ \\ \Rightarrow  \frac{x}{ \sqrt{ \frac{64}{900} } } = \frac{ \sqrt{ \frac{1024}{900} } }{  \sqrt{ \frac{4096}{10000} } } \Rightarrow  \frac{x}{ \frac{ \sqrt{64} }{ \sqrt{900} } } = \frac{ \frac{ \sqrt{1024} }{ \sqrt{900} } }{ \frac{ \sqrt{4096} }{ \sqrt{10000} } } \Rightarrow

\displaystyle  \frac{x}{ \frac{8}{30} } = \frac{ \frac{32}{30} }{ \frac{64}{100} } \Rightarrow x: \frac{8}{30} = \frac{32}{30} : \frac{64}{100} \Rightarrow x \cdot  \frac{30}{8} = \frac{32}{30} \cdot  \frac{100}{64} \Rightarrow  \\  \\  \\ \Rightarrow \frac{30x}{8} = \frac{3200}{1920} \Rightarrow 30x \cdot  \Rightarrow 1920=3200 \cdot 8 \Rightarrow 57600x=25600 \Rightarrow \\ \\ \Rightarrow x= \frac{25600}{57600} \Rightarrow \boxed{x= \frac{4}{9} }

\displaystyle 1).A= \sqrt{\left(2016- \frac{1008}{ \sqrt{1+3+5+7+...+2015} }\right)^{2015}:2015 }  \\  \\ 1+3+5+7+...+2015 \\ \\ 2015=1+(n-1) \cdot 2 \Rightarrow 2015=1+2n-2 \Rightarrow 2n=2015-1+2 \Rightarrow \\ \\ \Rightarrow 2n=2016 \Rightarrow n= \frac{2016}{2} \Rightarrow n=1008 \\ \\ S_{1008}= \frac{2 \cdot 1+(1008-1) \cdot 2}{2} \cdot 1008 \\ \\ S_{1008}=(2+1007 \cdot2) \cdot 504 \\ \\ S_{1008}=(2+2014) \cdot 504 \\ \\ S_{1008}=2016 \cdot 504 \\ \\ S_{1008}=1016064

\displaystyle 1+3+5+7+...+2015=1016064 \\ \\  \sqrt{1+3+5+7+...+2015} = \sqrt{1016064} =1008 \\ \\ A= \sqrt{\left(2016- \frac{1008}{1008} \right)^{2015}:2015} \\ \\ A=  \sqrt{(2016-1)^{2015}:2015} \\ \\ A= \sqrt{2015^{2015}:2015} \\ \\ A= \sqrt{2015^{2015-1}} \\ \\ A= \sqrt{2015^{2014}} \\ \\ A=2015^{ \frac{2014}{2} }\\ \\ \boxed{A=2015^{1007}}
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