Question

ajutorrrrrr ! Operatii cu numere reale !

Answer 1

$\displaystyle 1).(5+ \sqrt{2} )^2+(5- \sqrt{2} )^2= \\ =5^2+2 \cdot 5 \cdot \sqrt{2} +( \sqrt{2} )^2 +5^2-2 \cdot 5 \cdot \sqrt{2} +( \sqrt{2} )^2= \\ =25+10 \sqrt{2} +2+25-10 \sqrt{2} +2=54 \in N \\ 3). \sqrt{( \sqrt{3}-2)^2 } + \sqrt{(3- \sqrt{3} )^2} +|2 \sqrt{3} -3|= \\ =| \sqrt{3} -2|+|3- \sqrt{3} |+|2 \sqrt{3} -3|=2- \sqrt{3} +3- \sqrt{3} +2 \sqrt{3} -3=2$
[tex]\displaystyle 4). \sqrt{2} ,~ \sqrt[3]{3} ,~ \sqrt[6]{6} \\ c.m.m.m.c(2,3,6)=6 \\ \sqrt{2} = \sqrt[2]{2} = \sqrt[6]{2^3} = \sqrt[6]{8} \\ \sqrt[3]{3} = \sqrt[6]{3^2} = \sqrt[6]{9} \\ Crescator: \sqrt[6]{6} ,~ \sqrt[6]{8} ,~ \sqrt[6]{9} \Rightarrow \sqrt[6]{6} ,~ \sqrt{2} ,~ \sqrt[3]{3} [/tex]
 [tex]\displaystyle 5).3 \sqrt[3]{5} ,~4 \sqrt[3]{2} \\ (3 \sqrt[3]{5} )^3=27 \cdot 5=135 \\ (4 \sqrt[3]{2} )^3=64 \cdot 2=128 \\ 135\ \textgreater \ 128 \Rightarrow 3\sqrt[3]{5} \ \textgreater \ 4\sqrt[3]{2} [/tex]
$\displaystyle 6). \sqrt[3]{ \sqrt{64} } + \sqrt[3]{ \sqrt{729} } =5 \\ \sqrt[3]{ \sqrt{64} } + \sqrt[3]{ \sqrt{729} }= \sqrt[3]{8} + \sqrt[3]{27} = \sqrt[3]{2^3} + \sqrt[3]{3^3} =2+3=5 ~(A)$
$7).lg3=a,~lg900=2a+2 \\ lg900=lg30^2=2lg30=2(lg3+lg10)=2(a+1)=2a+2$
$8).a= \sqrt[4]{16} +log_381- \sqrt[3]{125} \\ a= \sqrt[4]{2^4} +log_33^4- \sqrt[3]{5^3} \\ a=2+4log_33-5 \\ a=2+4-5 \\ a=1 \in N$
$\displaystyle 12a).2^{-4}+1-4^{-2}= \frac{1}{2^4} +1- \frac{1}{4^2} = \frac{1}{16} +1- \frac{1}{16} =1 \\ b).\left( \frac{2}{3} \right)^{-1}- \sqrt[3]{ \frac{27}{8} } = \frac{3}{2} - \frac{ \sqrt[3]{27} }{ \sqrt[3]{8} } = \frac{3}{2} - \frac{ \sqrt[3]{3^3} }{ \sqrt[3]{2^3} }= \frac{3}{2}- \frac{3}{2} =0 \\ c).log_36+log_35-log_310=log_3(6 \cdot 5)-log_310= \\ =log_330-log_310=log_3 \frac{30}{10} =log_33=1$
$\displaystyle d).log_2 \frac{1}{2} +log_2 \frac{2}{3} +log_2 \frac{3}{4} +...+log_2 \frac{15}{16} = \\ =log_2 \left( \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot ... \cdot \frac{14}{15} \cdot \frac{15}{16} \right)=log_2 \frac{1}{16} = \\ =-log_216=-log_22^4=-4log_22=-4$
$\displaystyle e).(2+ \sqrt{3} )=2^2+2 \cdot 2 \cdot \sqrt{3} +( \sqrt{3} )^2=4+4 \sqrt{3} +3=4 \sqrt{3}+7 \\ \ [4 \sqrt{3}+7 ]=[4 \sqrt{3} ]+7= [\sqrt{16 \cdot 3}] = [\sqrt{48}]+7=6+7=13 \\ \ [4 \sqrt{3} +7]=[ \sqrt{48} +7]=13 \\ f). \frac{3}{4} + \frac{2}{5} + \frac{7}{10} = \frac{15+8+14}{20} = \frac{37}{20} =1,85 \\ \{1,85\}=1,85-[1,85]=1,85-1=0,85$