Question

пожайлуста помогите!!!

Answer 1

$\displaystyle 15.9)..(\frac{a}{6}- \frac{6}{a})*( \frac{1}{6+a}+ \frac{1}{6-a})= \frac{a^{2}-36}{6a}* \frac{6-a+6+a}{36-a^{2}}= \\ \\ \\ = -\frac{36-a^{2}}{6a}* \frac{12}{36-a^{2}}= -\frac{2}{a}; \\ \\ \\ 15.10)..( \frac{a-8}{a+8}- \frac{a+8}{a-8}): \frac{16a}{64-a^{2}}= \frac{(a-8)^{2}-(a+8)^{2}}{a^{2}-64}* \frac{64-a^{2}}{16a}= \\ \\ \\ =- \frac{(a-8+a+8)(a-8-a-8)}{64-a^{2}}* \frac{64-a^{2}}{16a}= \frac{2a*16}{16a}=2;$

$\displaystyle 15.11)..( \sqrt{7-4 \sqrt{3}}+ \sqrt{7+4 \sqrt{3}}})^{2}= \\ \\=( \sqrt{7-4 \sqrt{3}})^{2}+2* \sqrt{49-48}+( \sqrt{7+4 \sqrt{3}})^{2}= \\ \\ =7-4 \sqrt{3}+2+7+4 \sqrt{3}=16; \\ \\ \\ 15.12)..-3,25 \leq \frac{1-4x}{3}\ \textless \ 1,25 \\ \\-9,75 \leq 1-4x\ \textless \ 3,75 \\ \\1-4x \geq -9,75 \\ -4x \geq -10,75 \\ x \leq 2,6875 \\ \\ 1-4x\ \textless \ 3,75 \\ -4x\ \textless \ 2,75 \\ x\ \textgreater \ -0,6875$

При х∈Z:   x = {0; 1; 2}