Упростите рациональное алгебраическое выражение:
Требуется упростить рациональное выражение
\(\displaystyle \frac{b^2}{0{,}25xy}+4\frac{(b+x)^2}{x^2-xy}-4\frac{(b+y)^2}{xy-y^2}{\small .}\)
\(\displaystyle 0{,}25=\frac{1}{4}{\small ,}\) тогда
\(\displaystyle \frac{b^2}{0{,}25xy}=\frac{b^2}{\dfrac{1}{4}xy}= \frac{4b^2}{xy}\)
Разложим знаменатели дробей на множители:
\(\displaystyle \frac{4b^2}{xy}+4\frac{(b+x)^2}{x(x-y)}+4\frac{(b+y)^2}{y(x-y)}{\small .}\)
Ни одна из дробей не сокращается.
- \(\displaystyle \frac{4b^2}{xy}=\frac{4b^2(x-y)}{xy(x-y)}{ \small ,}\)
- \(\displaystyle 4\frac{(b+x)^2}{x(x-y)}=\frac{4(b+x)^2y}{xy(x-y)}{ \small ,}\)
- \(\displaystyle 4\frac{(b+y)^2}{y(x-y)}=\frac{4(b+y)^2x}{xy(x-y)}{\small .}\)
Записывая под одним знаменателем, получаем:
\(\displaystyle \frac{4b^2}{xy}+\frac{4(b+x)^2}{x^2-xy}-\frac{4(b+y)^2}{xy-y^2}=\)
\(\displaystyle =\frac{4b^2(x-y)+4(b+x)^2y-4(b+y)^2x}{xy(x-y)}=\)
\(\displaystyle 4\left(\frac{b^2(x-y)+(b+x)^2y-(b+y)^2x}{xy(x-y)}\right){\small .}\)
\(\displaystyle 4\left(\frac{b^2(x-y)+(b+x)^2y-(b+y)^2x}{xy(x-y)}\big)\right)\)
\(\displaystyle =4\left( \frac{ b^2x-b^2y+(b^2+2bx+x^2)y-(b^2+2by+y^2)x}{ xy(x-y) }\right)=\)
\(\displaystyle =4\left(\frac{ b^2x-b^2y+b^2y+2bxy+x^2y-b^2x-2bxy-xy^2}{ xy(x-y)}\right){\small .} \)
\(\displaystyle 4\left(\frac{ \cancel{\color{red}{ b^2x}}\color{green}{-}\cancel{\color{green}{ b^2y}}+\cancel{\color{green}{ b^2y}}+\cancel{\color{blue}{ 2bxy}}+x^2y\color{red}{-}\cancel{\color{red}{ b^2x}}\color{blue}{-}\cancel{\color{blue}{ 2bxy}}-xy^2}{ xy(x-y) }\right)=4\left(\frac{ x^2y-xy^2}{xy(x-y) }\right){\small .} \)
Разложим числитель на множители. Получаем:
\(\displaystyle 4\left(\frac{ x^2y-xy^2}{xy(x-y)}\right)=4\left(\frac{ xy(x-y)}{xy(x-y) }\right)=4\cdot1=4{\small .} \)
Ответ: \(\displaystyle 4{\small .} \)