Question

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

Answer 1

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

$-frac{p}{3} =-frac{p}{3}$

ЧТД

4.

$frac{9}{x} =frac{1-x}{x+4} +1$

$frac{9}{x} =frac{1-x}{x+4} +frac{x+4}{x+4}$

$frac{9}{x} =frac{5}{x+4}$

$9(x+4)=5x\9x+36=5x\4x+36=0\4x=-36\x=-9$

Answer 2

Отлично! Огромное спасибо @