Question

Buna! Ma puteti ajuta si pe mine?
Efectuati:
a) 24 + 30 - 42 + 36
    √3   5√3  7√3  √3
b)    16    +     12     -      6       -      8      
     2-√3        √3+1       2√3+1       3√3+5 
     

Answer 1

      
$a)\;\; \frac{24}{ \sqrt{3}} + \frac{30}{5\sqrt{3}}- \frac{42}{7\sqrt{3}}+\frac{36}{ \sqrt{3}} = \\ \\ = \frac{24*35}{35 \sqrt{3}} + \frac{30*7}{35\sqrt{3}}-\frac{42*5}{35\sqrt{3}}+\frac{36*35}{35\sqrt{3}} = \\ \\ = \frac{24*35+30*7-42*5+36*35}{35 \sqrt{3}}=\\ \\ = \frac{840+210-210+1260}{35 \sqrt{3}}= \\ \\ = \frac{840+1260}{35 \sqrt{3}}= \frac{2100}{35 \sqrt{3}}=\frac{60}{\sqrt{3}}=\frac{60\sqrt{3}}{3}=\frac{20\sqrt{3}}{1}=20\sqrt{3}$



$b)\;\; \frac{16}{2- \sqrt{3}}+\frac{12}{\sqrt{3}+1} - \frac{6}{2\sqrt{3}+1} - \frac{8}{3\sqrt{3}+5}= \;\;\;\;\;(\text{rationalizam numitorii}) \\ \\ =\frac{16(2+ \sqrt{3})}{4- 3}+\frac{12(\sqrt{3}-1)}{3-1} - \frac{6(2\sqrt{3}-1)}{12-1} - \frac{8(3\sqrt{3}-5)}{27-25}= \\ \\ =\frac{16(2+ \sqrt{3})}{1} +\frac{12(\sqrt{3}-1)}{2} - \frac{6(2\sqrt{3}-1)}{11} - \frac{8(3\sqrt{3}-5)}{2}=$


$=\frac{22*16(2+ \sqrt{3})}{22} +\frac{11*12(\sqrt{3}-1)}{22} - \frac{2*6(2\sqrt{3}-1)}{22} -\frac{11*8(3\sqrt{3}-5)}{22}= \\ \\ =\frac{352(2+ \sqrt{3})+132(\sqrt{3}-1)-12(2\sqrt{3}-1)- 88(3\sqrt{3}-5)}{22}= \\ \\ =\frac{352*2+ 352\sqrt{3}+132\sqrt{3}-132-24\sqrt{3}+12- 264\sqrt{3} +440)}{22}= \\ \\ =\frac{704-132+12+440+ (352 +132-24-264)\sqrt{3})}{22}= \\ \\ =\frac{1024+ 196\sqrt{3})}{22}=\frac{512+ 98\sqrt{3})}{11}$