Toán Lớp 8: Rút gọn biểu thức :
3x+1/(x-1)² – 1/x+1 + x+3/1-x²
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Toán Lớp 8: Rút gọn biểu thức :
3x+1/(x-1)² – 1/x+1 + x+3/1-x²
Comments ( 1 )
Giải đáp:
$\begin{array}{l}
\dfrac{{3x + 1}}{{{{\left( {x – 1} \right)}^2}}} – \dfrac{1}{{x + 1}} + \dfrac{{x + 3}}{{1 – {x^2}}}\\
= \dfrac{{\left( {3x + 1} \right).\left( {x + 1} \right) – {{\left( {x – 1} \right)}^2} – \left( {x + 3} \right)\left( {x – 1} \right)}}{{{{\left( {x – 1} \right)}^2}\left( {x + 1} \right)}}\\
= \dfrac{{3{x^2} + 4x + 1 – {x^2} + 2x – 1 – \left( {{x^2} + 2x – 3} \right)}}{{{{\left( {x – 1} \right)}^2}\left( {x + 1} \right)}}\\
= \dfrac{{2{x^2} + 6x – {x^2} – 2x + 3}}{{{{\left( {x – 1} \right)}^2}\left( {x + 1} \right)}}\\
= \dfrac{{{x^2} + 4x + 3}}{{{{\left( {x – 1} \right)}^2}\left( {x + 1} \right)}}\\
= \dfrac{{\left( {x + 1} \right)\left( {x + 3} \right)}}{{{{\left( {x – 1} \right)}^2}\left( {x + 1} \right)}}\\
= \dfrac{{x + 3}}{{{{\left( {x – 1} \right)}^2}}}
\end{array}$