Question

Помогите, дам 75 балов. виконайте тотожні перетворення раціонально виразу

Answer 1

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

Answer 2

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