Question

Упростите номер 34,35 и 36
Даю 30 баллов

Answer 1

$\displaystyle\bf\\34)\\\\1)\\\\m^{2} -\frac{1+m^{4} }{m^{2}-1 } =\frac{m^{2}\cdot(m^{2}-1)-(1+m^{4}) }{m^{2}-1 } =\\\\\\=\frac{m^{4}-m^{2}-1-m^{4} }{m^{2} -1}=-\frac{m^{2}+1 }{(m-1)\cdot(m+1)} \\\\\\2)\\\\-\frac{m^{2}+1 }{(m-1)\cdot(m+1)} :\frac{m^{2}+1 }{m-1} =-\frac{m^{2}+1 }{(m-1)\cdot(m+1)} \cdot\frac{m-1}{m^{2}+1 } =\\\\\\=-\frac{1}{m+1}$

$\displaystyle\bf\\35)\\\\1)\\\\\frac{1}{2-a} +\frac{2}{a^{2}-2a } =\frac{1}{2-a} +\frac{2}{a\cdot(a-2) } =\frac{1}{2-a} -\frac{2}{a\cdot(2-a) } =\\\\\\=\frac{a-2}{a\cdot(2-a)} =-\frac{1}{a}\\\\\\2)\\\\\frac{1}{a+1} :\Big(-\frac{1}{a}\Big)=-\frac{1}{a+1} \cdot\frac{a}{1} =-\frac{a}{a+1} \\\\\\3)\\\\\frac{a^{2}+1 }{a^{2}-1 }-\frac{a}{a+1}=\frac{a^{2} +1-a\cdot(a-1)}{a^{2}-1 } = \frac{a^{2} +1-a^{2} +a } {a^{2}-1} =\\\\\\=\frac{a+1}{(a+1)\cdot(a-1)} =\frac{1}{a-1}$

Задание под номером 36 такое же , как под номером 34 .