№ 7.14 Алгебра = № 12.14 Математика
Виконайте дії:
1. $\left(\frac{1}{1-a^2}-\frac{1}{a^2+2a+1}\right):\frac{2a}{a^2-1};$
2. $\left(\frac{x+1}{2x-2}-\frac{x+3}{2x+2}+\frac{6}{2x^2-2}\right)\cdot$
$\cdot\frac{4x^2-4}{5}.$
Розв'язок:
1. $\left(\frac{1}{1-a^2}-\frac{1}{a^2+2a+1}\right):\frac{2a}{a^2-1}=$
$= -\frac{1}{1+a}.$
а) $\frac{1}{1-a^2}-\frac{1}{a^2+2a+1}=$
$= \frac{1}{\left(1-a\right)\left(1+a\right)} -\frac{1}{\left(a+1\right)^2}=$
$= \frac{1+a-\left(1-a\right)}{\left(1+a\right)^2\left(1-a\right)}=$
$= \frac{1+a-1+a}{\left(1+a\right)^2\left(1-a\right)}=\frac{2a}{\left(1+a\right)^2\left(1-a\right)};$
б) $\frac{2a}{\left(1+a\right)^2\left(1-a\right)}:\frac{2a}{a^2-1}=$
$= \frac{2a\left(a-1\right)\left(a+1\right)}{\left(1+a\right)^2\left(1-a\right)\cdot2a}=$
$= \frac{a-1}{\left(1-a\right)\left(1+a\right)}=$
$= -\frac{1-a}{\left(1-a\right)\left(1+a\right)}=-\frac{1}{1+a}.$
2. $\left(\frac{x+1}{2x-2}-\frac{x+3}{2x+2}+\frac{6}{2x^2-2}\right)\cdot$
$\cdot\frac{4x^2-4}{5}=4.$
а) $\frac{x+1}{2x-2}-\frac{x+3}{2x+2}+\frac{6}{2x^2-2}=$
$= \frac{x+1}{2\left(x-1\right)}-\frac{x+3}{2\left(x+1\right)}+\frac{6}{2{(x}^2-1)}=$
$= \frac{\left(x+1\right)^2-\left(x+3\right)\left(x-1\right)+6}{2\left(x^2-1\right)}=$
$= \frac{x^2+2x+1-x^2+x-3x+3+6}{2\left(x^2-1\right)}=$
$= \frac{10}{2\left(x^2-1\right)}=\frac{5}{x^2-1};$
б) $\frac{5}{x^2-1}\cdot\frac{4\left(x^2-1\right)}{5}=\frac{5\cdot4\cdot\left(x^2-1\right)}{(x^2-1)\cdot5}=4.$