Задание 627

Решение:

• \({\largeа)}\ \frac{x^3-x^2+x+3}{x^2-2x+3}=\frac{ (x+1)(x^2-2x+3)}{x^2-2x+3}=x+1\)

• \(\begin{array}{l}\begin{array}{r}-\begin{array}{l}x^3-\phantom{0}x^2+\phantom{0}x+3\\x^3-2x^2+3x\\\hline\end{array}\begin{array}{|l}x^2-2x+3\\\hline{x}+1\end{array}\\\end{array}\\\phantom{\ 00000}\begin{array}{r}-\begin{array}{r}x^2-2x+3\\x^2-2x+3\\\hline\end{array}\end{array}\\\phantom{0000000000000000}\begin{array}{r}0\end{array}\end{array}\)

• \({\largeб)}\ \frac{x^3+x^2+3x-5}{x^2+2x+5}=\frac{ (x-1)(x^2+2x+5)}{x^2+2x+5}=x-1\)

• \(\begin{array}{l}\begin{array}{r}-\begin{array}{l}x^3+\phantom{0}x^2+3x-5\\x^3+2x^2+5x\\\hline\end{array}\begin{array}{|l}x^2+2x+5\\\hline{x}-1\end{array}\\\end{array}\\\phantom{0000}\begin{array}{r}-\begin{array}{r}{-}x^2-2x-5\\{-}x^2-2x-5\\\hline\end{array}\end{array}\\\phantom{0000000000000000}\begin{array}{r}0\end{array}\end{array}\)

• \({\largeв)}\ \frac{x^3-1}{x^3+2x^2+2x+1}=\frac{ (x-1)(x^2+x+1)}{x^3+2x^2+2x+1}=\frac{ (x-1)(x^2+x+1)}{ (x+1)(x^2+x+1)}=\frac{x-1}{x+1}\)

• \(\begin{array}{l}\begin{array}{r}-\begin{array}{l}x^3+2x^2+2x+1\\x^3+\phantom{0}x^2+\phantom{0}x\\\hline\end{array}\begin{array}{|l}x^2+x+1\\\hline{x}+1\end{array}\\\end{array}\\\phantom{\ 00000}\begin{array}{r}-\begin{array}{r}x^2+\phantom{0}x+1\\x^2+\phantom{0}x+1\\\hline\end{array}\end{array}\\\phantom{0000000000000000}\begin{array}{r}0\end{array}\end{array}\)

• \({\largeг)}\ \frac{x^3+8}{x^3-4x^2+8x-8}=\frac{ (x+2)(x^2-2x+4)}{x^3-4x^2+8x-8}=\frac{ (x+2)(x^2-2x+4)}{ (x-2)(x^2-2x+4)}=\frac{x+2}{x-2}\)

• \(\begin{array}{l}\begin{array}{r}-\begin{array}{l}x^3-4x^2+8x-8\\x^3-2x^2+4x\\\hline\end{array}\begin{array}{|l}x^2-2x+4\\\hline{x}-2\end{array}\\\end{array}\\\phantom{000}\begin{array}{r}-\begin{array}{r}{-}2x^2+4x-8\\{-}2x^2+4x-8\\\hline\end{array}\end{array}\\\phantom{0000000000000000}\begin{array}{r}0\end{array}\end{array}\)