№ 7.16 Алгебра = № 12.16 Математика
Виконайте дії:
1. $\left(2-\frac{2a^2-a}{a^2-a+1}\right):$
$: \left(\frac{1}{a+1}-\frac{a-1}{a^2-a+1}\right)= a+1;$
2. $\left(\frac{m-2}{m^2-2m+4}-\frac{6m-13}{m^3+8}\right)\cdot$
$\cdot\frac{2m^3+16}{18-6m}= \frac{3-m}{3}.$
Розв'язок:
1. а) $2-\frac{2a^2-a}{a^2-a+1}=\frac{2}{1}-\frac{2a^2-a}{a^2-a+1}=$
$= \frac{2{(a^2-a+1)-(2a}^2-a)}{a^2-a+1}=$
$= \frac{2{a^2-2a+2-2a}^2+a}{a^2-a+1}=\frac{2-a}{a^2-a+1};$
б) $\frac{1}{a+1}-\frac{a-1}{a^2-a+1}=$
$= \frac{\left(a^2-a+1\right)-\left(a-1\right)\left(a+1\right)}{\left(a+1\right)\left(a^2-a+1\right)}=$
$= \frac{a^2-a+1-a^2+1}{\left(a+1\right)\left(a^2-a+1\right)}=$
$= \frac{2-a}{\left(a+1\right)\left(a^2-a+1\right)};$
в) $\frac{2-a}{a^2-a+1}:\frac{2-a}{\left(a+1\right)\left(a^2-a+1\right)}=$
$= \frac{2-a}{a^2-a+1}\cdot\frac{\left(a+1\right)\left(a^2-a+1\right)}{2-a}=$
$= \frac{\left(2-a\right)\left(a+1\right){(a}^2-a+1)}{{(a}^2-a+1)\left(2-a\right)}=a+1.$
2. а) $\frac{m-2}{m^2-2m+4}-\frac{6m-13}{m^3+8}=$
$= \frac{m-2}{m^2-2m+4}-$
$- \frac{6m-13}{\left(m+2\right)\left(m^2-2m+4\right)}=$
$= \frac{\left(m-2\right)\left(m+2\right)-\left(6m-13\right)}{\left(m+2\right)\left(m^2-2m+4\right)}=$
$= \frac{m^2-4-6m+13}{\left(m+2\right)\left(m^2-2m+4\right)}=$
$= \frac{m^2-6m+9}{\left(m+2\right)\left(m^2-2m+4\right)}=\frac{\left(m-3\right)^2}{m^3+8};$
б) $\frac{\left(m-3\right)^2}{m^3+8}\cdot\frac{2m^3+16}{18-6m}=$
$= \frac{\left(m-3\right)^2\cdot2\left(m^3+8\right)}{{(m}^3+8)\cdot6\left(3-m\right)}=$
$= \frac{\left(3-m\right)^2\cdot2\left(m^3+8\right)}{\left(3-m\right)\cdot6\left(m^3+8\right)}=\frac{3-m}{3}.$