a)A=(2x+1/2x-1-2x-1/2x+1) : 4x/10-5 b)B=(1/x^2+x-2-x/x+1) :(1/x+x-2)

1 bình luận về “a)A=(2x+1/2x-1-2x-1/2x+1) : 4x/10-5 b)B=(1/x^2+x-2-x/x+1) :(1/x+x-2)”

1. Giải đáp:$\begin{array}{l}
   a)A = \dfrac{{10}}{{2x + 1}}\\
   b)B = \dfrac{1}{{x + 1}}
   \end{array}$

    

   Lời giải và giải thích chi tiết:

   $\begin{array}{l}
   Dkxd:x \ne \dfrac{1}{2};x \ne  – \dfrac{1}{2};x \ne 0\\
   A = \left( {\dfrac{{2x + 1}}{{2x – 1}} – \dfrac{{2x – 1}}{{2x + 1}}} \right):\dfrac{{4x}}{{10x – 5}}\\
    = \dfrac{{{{\left( {2x + 1} \right)}^2} – {{\left( {2x – 1} \right)}^2}}}{{\left( {2x – 1} \right)\left( {2x + 1} \right)}}.\dfrac{{5.\left( {2x – 1} \right)}}{{4x}}\\
    = \dfrac{{4{x^2} + 4x + 1 – 4{x^2} + 4x – 1}}{{2x + 1}}.\dfrac{5}{{4x}}\\
    = \dfrac{{8x}}{{2x + 1}}.\dfrac{5}{{4x}}\\
    = \dfrac{{10}}{{2x + 1}}\\
   b)Dkxd:x \ne 0;x \ne  – 1\\
   B = \left( {\dfrac{1}{{{x^2} + x}} – \dfrac{{2 – x}}{{x + 1}}} \right):\left( {\dfrac{1}{x} + x – 2} \right)\\
    = \left( {\dfrac{1}{{x\left( {x + 1} \right)}} – \dfrac{{2 – x}}{{x + 1}}} \right):\dfrac{{1 + x\left( {x – 2} \right)}}{x}\\
    = \dfrac{{1 – x.\left( {2 – x} \right)}}{{x\left( {x + 1} \right)}}.\dfrac{x}{{1 + {x^2} – 2x}}\\
    = \dfrac{{1 – 2x + {x^2}}}{{x + 1}}.\dfrac{1}{{{x^2} – 2x + 1}}\\
    = \dfrac{1}{{x + 1}}
   \end{array}$

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