Question

15. спростіть вираз ​

Answer 1

$\left ( \frac{a+1}{a-1}-\frac{4a}{a^2-1} \right ):\frac{a-1}{a^2+a}=\left ( \frac{a+1}{a-1}-\frac{4a}{(a-1)(a+1)} \right )\cdot \frac{a(a+1)}{a-1}$$\frac{a+1}{a-1}\cdot \frac{a(a+1)}{a-1}-\frac{4a}{(a-1)(a+1)}\cdot \frac{a(a+1)}{a-1}=\frac{a(a+1)^2}{(a-1)^2}-\frac{4a^2}{(a-1)^2}$$\frac{a\left ( a^2+2a+1 \right )-4a^2}{(a-1)^2}=\frac{a^3+2a^2+a-4a^2}{(a-1)^2}=\frac{a^3-2a^2+a}{(a-1)^2}$$\frac{a\left ( a^2-2a+1 \right )}{(a-1)^2}=\frac{a(a-1)^2}{(a-1)^2}=a, \; a\neq 1$

Answer 2

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