Môn Toán Lớp 8: Thực hiện phép tính X^2+5x/x^2 – 2x+1-5x+1/x^2-2x+1 5x+10/4x-8÷2x+4/4-2x

Giải đáp:

$\begin{array}{l}
Dkxd:{x^2} – 2x + 1 \ne 0\\
 \Leftrightarrow {\left( {x – 1} \right)^2} \ne 0\\
 \Leftrightarrow x – 1 \ne 0\\
 \Leftrightarrow x \ne 1
\end{array}$

$\begin{array}{l}
a)\dfrac{{{x^2} + 5x}}{{{x^2} – 2x + 1}} – \dfrac{{5x + 1}}{{{x^2} – 2x + 1}}\\
 = \dfrac{{{x^2} + 5x – 5x – 1}}{{{x^2} – 2x + 1}}\\
 = \dfrac{{{x^2} – 1}}{{{{\left( {x – 1} \right)}^2}}}\\
 = \dfrac{{\left( {x – 1} \right)\left( {x + 1} \right)}}{{{{\left( {x – 1} \right)}^2}}}\\
 = \dfrac{{x + 1}}{{x – 1}}\\
b)\dfrac{{5x + 10}}{{4x – 8}}:\dfrac{{2x + 4}}{{4 – 2x}}\\
 = \dfrac{{5.\left( {x + 2} \right)}}{{4.\left( {x – 2} \right)}}.\dfrac{{2.\left( {2 – x} \right)}}{{2.\left( {x + 2} \right)}}\\
 = \dfrac{5}{4}.\left( { – 1} \right)\\
 =  – \dfrac{5}{4}
\end{array}$

$Dkxd:\left\{ \begin{array}{l}
4x – 8 \ne 0\\
4 – 2x \ne 0\\
2x + 4 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne 2\\
x \ne  – 2
\end{array} \right.$