Question

Teza 1 partea a 2 ex 1 punctele a,b, si c dau coroana.

Answer 1

[tex]a) \sqrt{48} -( \frac{16}{ \sqrt{32} }- \frac{18}{ \sqrt{243} } )* \sqrt{6} = \sqrt{16*3}- ( \frac{16}{ \sqrt{16*2} }- \frac{18}{ \sqrt{81*3} } )* \sqrt{6}= \\ =4 \sqrt{3}- ( \frac{16}{ 4\sqrt{2} }- \frac{18}{ 9\sqrt{3} } )* \sqrt{6} =4 \sqrt{3}- ( 4\sqrt{2} }-2\sqrt{3} })* \sqrt{6}= \\ =4 \sqrt{3}-4 \sqrt{12}+2 \sqrt{18}=4 \sqrt{3}-4 \sqrt{4*3}+2 \sqrt{9*2}= \\ =4 \sqrt{3}-8\sqrt{3}+6 \sqrt{2}=6 \sqrt{2}-4 \sqrt{3} [/tex]
[tex]b) a= \frac{ \sqrt{18}+9}{ \sqrt{128} +24}= \frac{ \sqrt{9*2}+9 }{ \sqrt{64*2} +24} = \frac{3 \sqrt{2}+9 }{8 \sqrt{2}+24} \\ b= \frac{ \sqrt{20}-4 }{ \sqrt{45}-6 }= \frac{ \sqrt{4*5} -4}{ \sqrt{9*5}-6 } = \frac{2 \sqrt{5}-4 }{3 \sqrt{5}-6 } \\ 4ab-1=4*\frac{3 \sqrt{2}+9 }{8 \sqrt{2}+24}* \frac{2 \sqrt{5}-4 }{3 \sqrt{5}-6 }-1= \frac{4(3 \sqrt{2}+9)(2 \sqrt{5}-4 )}{ (8\sqrt{2}+24)(3 \sqrt{5}-6 ) } -1= \\ = \frac{4(6 \sqrt{10}-12 \sqrt{2}+18 \sqrt{5}-36)}{24 \sqrt{10}-48 \sqrt{2}+72 \sqrt{5} -144 }-1 = [/tex]
$= \frac{24 \sqrt{10}-48 \sqrt{2}+72 \sqrt{5}-144)}{24 \sqrt{10}-48 \sqrt{2}+72 \sqrt{5} -144 }-1=1-1=0$