Question

Помогите пожалуйста с 11 и 12 заданиями

Answer 1

$2x; 4y; 8z; 16w\\x=8;w=27\\\\2x=2*8=16\\16w=16*27=432\\\\b_1=16\\b_2=4y\\b_3=8z\\b_4=432\\\\q=b_2/b_1=4y/16=y/4\\q=b_3/b_2=8z/4y=2z/y\\q=b_4/b_3=432/8z=54/z\\\\ \frac{y}{4}= \frac{2z}{y}=\ \textgreater \ 8z=y^2=\ \textgreater \ z= \frac{y^2}{8}\\\\ \frac{y}{4}= \frac{54}{z} =\ \textgreater \ z= \frac{54*4}{y}= \frac{216}{y} \\\\ \frac{y^2}{8}= \frac{216}{y}=\ \textgreater \ y^3=1728=\ \textgreater \ y= \sqrt[3]{1728}=12\\\\z= \frac{y^2}{8}= \frac{12^2}{8}= \frac{144}{8}=18\\\\z=18$


$( \frac{1}{4a^2}- \frac{16}{b^2} )* \frac{2ab}{8a-b}= \frac{b^2-64a^2}{4a^2b^2}* \frac{2ab}{8a-b}= \frac{(b-8a)(b+8a)*2ab}{-4a^2b^2(b-8a)}=\\\\= -\frac{b+8a}{2ab}\\\\a=2- \sqrt{3};\; \; \; b=16+8 \sqrt{3}\\\\ -\frac{16+8 \sqrt{3}+8(2- \sqrt{3}) }{2(2- \sqrt{3})(16+8 \sqrt{3}) }=- \frac{16+8 \sqrt{3}+16-8 \sqrt{3} }{2*8(2- \sqrt{3})(2+ \sqrt{3}) }=- \frac{32}{16(4-3)}=- \frac{32}{16}=-2$