Question

Решите пожалуйста со второго по четвертое задание

Answer 1

№2

a)     $\frac{18a^{15}b^{4}}{54a^{5}b^{8}} = \frac{a^{10}}{3b^{4}}$         б)    $\frac{a^{2}-b^{2}}{5a-5b} = \frac{(a-b)(a+b)}{5(a-b)} = \frac{a+b}{5}$

в)    $\frac{x^{2}+12x+36}{x^{2}-36} = \frac{(x+6)^{2}}{(x-6)(x+6)} = \frac{x+6}{x-6}$

№3

а) $\frac{3b-2}{4b^{3}}-\frac{6b^{2}-5}{10b^{4}} =\frac{5b(3b-2)-2(6b^{2}-5)}{20b^{4}} = \frac{15b^{2}-10b-12b^{2}+10}{20b^{4} }=\frac{3b^{2}-10b+10}{20b^{4}}$

б) $\frac{1}{x+5}+\frac{4x}{x^{2}-25}=\frac{1}{x+5}+\frac{4x}{(x-5)(x+5)}=\frac{1}{x+5}(1+\frac{4x}{x-5})=\frac{1}{x+5}(\frac{x-5+4x}{x-5})=\frac{1}{x+5}(\frac{5x-5}{x-5})=\frac{5(x-1)}{(x+5)(x-5)}=\frac{5(x-1)}{(x-5)^{2} }$

в)  $\frac{4}{x^{2}-9}-\frac{2}{x^{2}+3x}=\frac{4}{(x-3)(x+3)}-\frac{2}{x(x+3)}=\frac{2}{x+3}(\frac{2}{x-3}-\frac{1}{x})=\frac{2}{x+3}(\frac{2x-(x-3)}{x(x-3)})=\frac{2}{x+3}(\frac{2x-x+3}{x(x-3)})=\frac{2}{x+3}(\frac{x+3}{x(x-3)})=\frac{2}{x(x-3)}=\frac{2}{x^{2}-3x}$

г)  $\frac{1-3a^{2}}{a+2}+3a=\frac{1-3a^{2}+3a(a+2)}{a+2}=\frac{1-3a^{2}+3a^{2}+6a}{a+2}=\frac{1+6a}{a+2}$

№4

а)  $\frac{n+3}{2n+2}-\frac{n+1}{2n-2}+\frac{3}{n^{2}-1}=\frac{n+3}{2(n+1)}-\frac{n+1}{2(n-1)}+\frac{3}{(n-1)(n+1)}=\frac{(n+3)(n-1)-(n+1)(n+1)+2*3}{2(n-1)(n+1)}=\frac{n^{2}-n+3n-3-(n^{2}+n+n+1)+6}{2(n-1)(n+1)}=\frac{n^{2}+2n-3-n^{2}-2n-1+6}{2(n-1)(n+1)}=\frac{2}{2(n-1)(n+1)}=\frac{1}{n^{2}-1}$

б) $\frac{2a^{2} +7a+9}{a^{3}-1}+\frac{4a+3}{a^{2}+a+1}=\frac{2a^{2} +7a+9+(4a+3)(a-1)}{(a-1)(a^{2}+a+1)}=\frac{2a^{2} +7a+9+4a^{2}-4a+3a-3}{(a-1)(a^{2}+a+1)}=\frac{6a^{2}+6a+6}{(a-1)(a^{2}+a+1)}=\frac{6(a^{2}+a+1)}{(a-1)(a^{2}+a+1)}=\frac{6}{a-1}$