Question

плиз помогите пожалуйста дам все что эсть побистрее​

Answer 1

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

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