Verified answer

[tex]\displaystyle\bf\\1)\\\\\frac{1}{b^{2} +3b} -\frac{1}{b+3} +1=\frac{1}{b\cdot(b +3)} -\frac{1}{b+3} +1=\\\\\\=\frac{1-b+b^{2}+3b }{b(b+3)} =\frac{b^{2} +2b+1}{b(b+3)} =\frac{(b+1)^{2} }{b(b+3)} \\\\\\2)\\\\\frac{1}{b^{2}-1 } +\frac{1}{(b+1)^{3} } =\frac{1}{(b+1 )(b-1)} +\frac{1}{(b+1)^{3} } =\\\\\\=\frac{(b+1)^{2}+b-1 }{(b-1)(b+1)^{3} } =\frac{b^{2}+2b+1+b-1 }{(b-1)(b+1)^{3} } =\frac{b^{2}+3b }{(b-1)(b+1)^{3} } =\\\\\\=\frac{b(b+3)}{(b-1)(b+1)^{3} }[/tex]

[tex]\displaystyle\bf\\3)\\\\\frac{(b+1)^{2} }{b(b+3)} \cdot\frac{b(b+3)}{(b-1)(b+1)^{3} } =\frac{1}{(b-1)(b+1)} =\frac{1}{b^{2} -1} \\\\\\4)\\\\\frac{1}{b^{2}-1 } -\frac{b^{2} }{b^{2} -1} =\frac{1-b^{2} }{b^{2} -1} =-\frac{b^{2}-1 }{b^{2} -1} =-1\\\\\\-1=-1[/tex]

Что и требовалось доказать .