Question

Rezolvare acestui exercitiu

Answer 1

 2)  
$(\frac{ 2^{3} a^{-2} b^{-1}}{24ab^{-2}})^{-1}= \frac{ a^{3} }{3b} \\ \\ \frac{ 2^{-3} a^{2}b^{1}}{24^{-1}a^{-1}b^{2}}= \frac{ a^{3} }{3b} \\ \\ \frac{24a^{2-(-1)} }{ 2^{3} b^{2-1}}= \frac{ a^{3} }{3b} \\ \\ \frac{24a^{3}}{8b}= \frac{ a^{3} }{3b} \\ \\ \frac{3a^{3}}{b}= \frac{ a^{3} }{3b} \,\,\,FALS$


5a)
$\sqrt{48}-13 \sqrt{ \frac{12}{25}}+ \sqrt{4 \frac{8}{25}}= \\ \\ = \sqrt{16*3}-13 \sqrt{ \frac{4*3}{25}}+ \sqrt{\frac{4*25+8}{25}}= \\ \\ =4\sqrt{3}-\frac{13*2}{5} \sqrt{3}+ \sqrt{\frac{108}{25}}= \\ \\ = 4\sqrt{3}-\frac{13*2}{5} \sqrt{3}+ \sqrt{\frac{36*3}{25}}= \\ \\ = 4\sqrt{3}-\frac{13*2}{5} \sqrt{3}+ \frac{6}{5} \sqrt{3}= \\ \\ =(4-\frac{13*2}{5}+\frac{6}{5})*\sqrt{3}= \\ \\=( \frac{20}{5} -\frac{13*2}{5}+\frac{6}{5})*\sqrt{3}= \frac{20-26+6}{5}*\sqrt{3}= \frac{0}{5}*\sqrt{3}=0$


5b)
$( \sqrt{11+6 \sqrt{2}}-\sqrt{11-6 \sqrt{2}})^{2}= \\=(\sqrt{11+6 \sqrt{2}})^{2}-2( \sqrt{11+6 \sqrt{2}})(\sqrt{11-6 \sqrt{2}})+ (\sqrt{11-6 \sqrt{2}})^{2}= \\ =11+6 \sqrt{2} -2(\sqrt{(11+6 \sqrt{2})(11-6 \sqrt{2})}+11-6 \sqrt{2}= \\ =11+6 \sqrt{2}+11-6 \sqrt{2}-2 \sqrt{11^{2}-(6 \sqrt{2})^{2}}= \\ =22-2\sqrt{121-(36 *2}= \\ 22 -2 \sqrt{121-72} = \\ =22-2 \sqrt{49}= \\ =22-2*7= \\ =22-14=8$