Задание 20 стр 106

Question

Выполните действия.
ewline
a) $\frac{3}{x^2}+\frac{5}{x} \mid$ b) $\frac{7}{6 a^2}+\frac{5}{3 a}\mid$ c) $\frac{x^2-1}{x+1}-\frac{x^2+1}{x-1}$
d) $\frac{18}{y^2-9}+\frac{3}{3-y}$
ewline e) $\frac{3 a+3}{6-3 a}-\frac{a+2}{a^2-4} \mid$ f) $\frac{1-x}{x^2+2 x+1}+\frac{1}{x+1} \mid$ g) $\frac{x+1}{(x-1)^2}+\frac{x+4}{x^2+3 x-4}$

Gimli · Accepted Answer

\begin{enumerate}
\item $\frac{3}{x^2}+\frac{5}{x}=\frac{3}{x^2}+\frac{5 x}{x^2}=\frac{3+5 x}{x^2}$;
\item $\frac{7}{6 a^2}+\frac{5}{3 a}=\frac{7}{6 a^2}+\frac{5 \cdot 2 a}{6 a^2}=\frac{7+10 a}{6 a^2}$
\item $\frac{x^2-1}{x+1}-\frac{x^2+1}{x-1}=\frac{(x-1)(x+1)}{x+1}-\frac{x^2+1}{x-1}=x-1-\frac{x^2+1}{x-1} = \frac{(x-1)^2}{x-1}-\frac{x^2+1}{x-1} = \frac{x^2-2x+1-x^2-1}{x-1}=\frac{-2x}{x-1}.$
\item $\frac{18}{y^2-9}+\frac{3}{3-y}=-\frac{18}{9-y^2}+\frac{3}{3-y}=-\frac{18}{9-y^2}+\frac{3(3+y)}{9-y^2}=\frac{-18+9+3y}{9-y^2}=\frac{-9+3y}{9-y^2}=\frac{3(y-3)}{(3-y)(3+y)} = -\frac{3}{3+y}.$
\item $\frac{3 a+3}{6-3 a}-\frac{a+2}{a^2-4}=\frac{3(a+1)}{3(2-a)}+\frac{a+2}{4-a^2}=\frac{a+1}{2-a}+\frac{2+a}{(2+a)(2-a)}=\frac{a+1}{2-a}+\frac{1}{2-a}=\frac{a+1+1}{2-a}=\frac{2+a}{2-a}$.
\item $\frac{1-x}{x^2+2 x+1}+\frac{1}{x+1}=\frac{1-x}{(x+1)^2}+\frac{x+1}{(x+1)^2}=\frac{1-x+x+1}{(x+1)^2}=\frac{2}{(x+1)^2}$.
\item $\frac{x+1}{(x-1)^2}+\frac{x+4}{x^2-x+4 x-4}=\frac{x+1}{(x-1)^2}+\frac{x+4}{x(x-1)+4(x-1)}=$ $=\frac{x+1}{(x-1)^2}+\frac{x+4}{(x+4)(x-1)}=\frac{x+1}{(x-1)^2}+\frac{1}{x-1}=\frac{x+1}{(x-1)^2}+\frac{x-1}{(x-1)^2}=$ $=\frac{x+1+x-1}{(x-1)^2}=\frac{2 x}{(x-1)^2}$
\end{enumerate}

Задание 20 стр 106

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Задание 20 стр 106

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