№ 7.29 Алгебра = № 12.29 Математика
Доведіть, що значення виразу
$\left(\frac{m^2-3m}{m^3+3m^2+3m+1}+\frac{1}{m^2+2m+1}\right)\cdot$
$\cdot\left(\frac{3-m}{m^2-2m+1}-\frac{2}{1-m}\right)$ є додатним для всіх допустимих значень змінної.
Розв'язок:
$\left(\frac{m^2-3m}{m^3+3m^2+3m+1}+\frac{1}{m^2+2m+1}\right)\cdot$
$\cdot\left(\frac{3-m}{m^2-2m+1}-\frac{2}{1-m}\right)=$
$= \frac{1}{\left(m+1\right)^2}>0;$
а) $\frac{m^2-3m}{m^3+3m^2+3m+1}+\frac{1}{m^2+2m+1}=$
$= \frac{m^2-3m}{{(m}^3+1)+(3m^2+3m)}+\frac{1}{\left(m+1\right)^2}=$
$= \frac{m^2-3m}{\left(m+1\right)\left(m^2-m+1\right)+3m\left(m+1\right)}+$
$+ \frac{1}{\left(m+1\right)^2}=$
$=\frac{m^2-3m}{\left(m+1\right)\left(m^2-m+1+3m\right)}+$
$+ \frac{1}{\left(m+1\right)^2}= \frac{m^2-3m}{\left(m+1\right)\left(m^2+2m+1\right)}+$
$+ \frac{1}{\left(m+1\right)^2}=$
$= \frac{m^2-3m}{\left(m+1\right)\left(m+1\right)^2}+\frac{1}{\left(m+1\right)^2}=$
$= \frac{m^2-3m}{\left(m+1\right)^3}+\frac{1}{\left(m+1\right)^3}=$
$= \frac{\left(m^2-3m\right)+\left(m+1\right)}{\left(m+1\right)^3}= $
$= \frac{m^2-3m+m+1}{\left(m+1\right)^3}=$
$= \frac{m^2-2m+1}{\left(m+1\right)^3}=\frac{\left(m-1\right)^2}{\left(m+1\right)^3};$
б) $\frac{3-m}{m^2-2m+1}-\frac{2}{1-m}=$
$= \frac{3-m}{\left(m-1\right)^2}+\frac{1}{\left(m-1\right)}=$
$= \frac{\left(3-m\right)+2\left(m-1\right)}{\left(m+1\right)^2}=$
$= \frac{3-m+2m-2}{\left(m+1\right)^2}=\frac{m+1}{\left(m+1\right)^2};$
в) $\frac{\left(m-1\right)^2}{\left(m+1\right)^3}\cdot\frac{(m+1)}{\left(m+1\right)^2}=$
$= \frac{(m+1)\cdot\left(m-1\right)^2}{\left(m+1\right)^3\cdot\left(m-1\right)^2}=\frac{1}{\left(m+1\right)^2}>0.$