Нужна помощь в решении сам. работы по алгебре 8 класс. Фотки заданий:

Question 1

Question 2

1.

а) $\frac{18p^{3}}{k^{5}} *\frac{k^{6}}{24p^{9}} =\frac{18p^{3}*k^{6}}{k^{5}*24p^{9}} =\frac{3k}{4p^{6}}$

б) $\frac{5a^{8}}{3+a} :\frac{15a^{4}}{a^{2}+6a+9} =\frac{5a^{8}}{3+a} :\frac{15a^{4}}{(a+3)^{2}} =\frac{5a^{8}*(a+3)^{2}}{(3+a)*15a^{4}} =\frac{a^{4}*(a+3)}{3} =\frac{a^{5}+3a^{4}}{3}$

в) $\frac{4y^{2}-1}{y^{2}-9} :\frac{6y+3}{y+3} =\frac{(2y-1)(2y+1)}{(y-3)(y+3)} :\frac{3(2y+1)}{y+3}= \frac{(2y-1)(2y+1)(y+3)}{(y-3)(y+3)*3(2y+1)} =\frac{2y-1}{3(y-3)}=\frac{2y-1}{3y-9}$

2.

$(\frac{x}{x-3} -\frac{2}{x+3} ):\frac{4x^{2}+4x+24}{x^{2}-9} =(\frac{x(x+3)}{x^{2}-9} -\frac{2(x-3)}{x^{2}-9})*\frac{x^{2}-9}{4(x^{2}+x+6)} =\frac{x^{2}+3x-2x+6}{x^{2}-9} *\frac{x^{2}-9}{4(x^{2}+x+6)} =\frac{x^{2}+x+6}{x^{2}-9} *\frac{x^{2}-9}{4(x^{2}+x+6)} =\frac{1}{4} =0.25$

3. $\frac{9-p^{2}}{3p+9} *\frac{p^{2}}{(3-p)^{2}} +\frac{p}{p-3} =\frac{(3-p)(3+p)}{3(p+3)} *\frac{p^{2}}{(3-p)^{2}} +\frac{p}{p-3} =\frac{p^{2}}{3(3-p)} +\frac{p}{p-3} =\frac{p^{2}}{3(3-p)} -\frac{3p}{3(3-p)} =\frac{p^{2}-3p}{3(3-p)} =\frac{-p(3-p)}{3(3-p)}=-\frac{p}{3}$