Question

Упростите помогите пжпжпжпжрж​

Answer 1

$\displaystyle\bf\\1)\\\\\frac{a}{a-2} -\frac{a^{2}+3a }{a^{2}-4 } =\frac{a}{a-2} -\frac{a^{2}+3a }{(a-2)(a+2) } =\frac{a\cdot(a+2)-a^{2}-3a }{(a-2)(a+2)} =\\\\\\=\frac{a^{2} +2a-a^{2} -3a}{(a-2)(a+2)} =\frac{-a}{a^{2} -4} =\frac{a}{4-a^{2} } \\\\\\2)\\\\\frac{2a^{2}+a+10 }{4a^{2}-25 } +\frac{a}{5-2a} =\frac{2a^{2} +a+10}{(2a-5)(2a+5)} -\frac{a}{2a-5} =\\\\\\=\frac{2a^{2}+a+10-a\cdot(2a+5) }{(2a-5)(2a+5)} =\frac{2a^{2}+a+10-2a^{2} -5a }{(2a-5)(2a+5)} =$

$\displaystyle\bf\\=\frac{10-4a}{(2a-5)(2a+5)} =\frac{2(5-2a)}{(2a-5)(2a+5)} =-\frac{2}{2a+5} \\\\\\3)\\\\\frac{b+2}{b+3} -\frac{b^{2}-20 }{b^{2}-9 } =\frac{b+2}{b+3} -\frac{b^{2}-20 }{(b-3)(b+3) } =\frac{(b+2)(b-3)-b^{2} +20}{(b-3)(b+3)} =\\\\\\=\frac{b^{2}-3b+2b-6-b^{2} +20 }{(b-3)(b+3)} =\frac{14-b}{b^{2} -9} \\\\\\4)\\\\\frac{2b}{4-3b} +\frac{6b^{2}+2b-8 }{9b^{2} -16} =\frac{2b}{4-3b} -\frac{6b^{2}+2b-8 }{(4-3b)(3b+4)} =$

$\displaystyle\bf\\=\frac{2b\cdot(3b+4)-6b^{2} -2b+8}{(4-3b)(4+3b)} =\frac{6b^{2} +8b-6b^{2} -2b+8}{(4-3b)(4+3b)} =\\\\\\=\frac{6b+8}{(4-3b)(4+3b)} =\frac{2(3b+4)}{(4-3b)(4+3b)}=\frac{2}{4-3b}$