Help dam naj zad w załączniku

Help Dam Naj Zad W Załączniku class=

Odpowiedź :

a)

[tex]\frac{|\sqrt{27}-4 |}{|1-\sqrt{3} |} -3*|4-2\sqrt{3} |=\frac{|3\sqrt{3}-4 |}{\sqrt{3}-1 } -3(4-2\sqrt{3})=\frac{3\sqrt{3}-4 }{\sqrt{3}-1 } -12+6\sqrt{3}=\frac{(3\sqrt{3}-4)*(\sqrt{3}+1) }{2}-12+6\sqrt{3}=\frac{9+3\sqrt{3}-4\sqrt{3}-4 }{2}-12+6\sqrt{3}=\frac{5-\sqrt{3} }{2}-12+6\sqrt{3}=\frac{5-\sqrt{3}-24+12\sqrt{3} }{2}=\frac{-19+11\sqrt{3} }{2}[/tex]

b)

[tex]|(\sqrt{6}-\sqrt{5})(\sqrt{5}+\sqrt{6})|-3(\sqrt{20}-2*|\sqrt{5}-2|)=|(\sqrt{6}-\sqrt{5})(\sqrt{6}+\sqrt{5})|-3(2\sqrt{5}-2(\sqrt{5}-2))= |6-5|-3(2\sqrt{5}-2\sqrt{5}+4)=|6-5|-3*4=|1|-12=1-12=-11[/tex]