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2 câu trả lời 431
\[\begin{array}{l}
E = \left( {\frac{{x + 2}}{{x - 1}} - \frac{{x - 2}}{{x + 1}}} \right).\frac{{{x^2} - 1}}{{x + 2}}\\
dk:x \ne \pm 1;x \ne - 2\\
= \frac{{\left( {x + 2} \right)\left( {x + 1} \right) - \left( {x - 2} \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}.\frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{x + 2}}\\
= \frac{{{x^2} + x + 2{\rm{x}} + 2 - {x^2} + x + 2{\rm{x}} - 2}}{{x + 2}}\\
= \frac{{6{\rm{x}}}}{{x + 2}}\\
Vay:E = \frac{{6{\rm{x}}}}{{x + 2}}\left( {x \ne \pm 1;x \ne - 2} \right)
\end{array}\]
DkXD: x≠±1, x≠-2
E=[(x+2)/(x-1)-(x-2)/(x+1)]×(x²-1)/(x+2
E=[(x+2)(x+1)/(x+1)(x-1)-(x-2)(x-1)/(x+1)(x-1)]×(x+1)(x-1)/(x+2)
E=[(x+2)(x+1)-(x-2)(x-1)/(x+1)(x-1)]×(x+1)(x-1)/(x+2)
E=(x²+x+2x+2-x²+x+2x-2)/(x+1)(x-1)×(x+1)(x-1)/(x+2)
E=6x/(x+1)(x-1)×(x+1)(x-1)/(x+2)
E=6x/(x+2)
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