tính A=(x/x^2-25-x-5/x^2+ 5 x): (2 x-5/x^2+ 5 x) +x/5-x

0 bình luận về “tính A=(x/x^2-25-x-5/x^2+ 5 x): (2 x-5/x^2+ 5 x) +x/5-x”

1. Đáp án:

   \[ – 1\]

   Giải thích các bước giải:

   \(\begin{array}{l}
   A = \left( {\frac{x}{{{x^2} – 25}} – \frac{{x – 5}}{{{x^2} + 5x}}} \right):\left( {\frac{{2x – 5}}{{{x^2} + 5x}}} \right) + \frac{x}{{5 – x}}\\
    = \left( {\frac{x}{{\left( {x – 5} \right)\left( {x + 5} \right)}} – \frac{{\left( {x – 5} \right)}}{{x\left( {x + 5} \right)}}} \right):\left( {\frac{{2x – 5}}{{x\left( {x + 5} \right)}}} \right) + \frac{x}{{5 – x}}\\
    = \left( {\frac{{{x^2} – {{\left( {x – 5} \right)}^2}}}{{x\left( {x – 5} \right)\left( {x + 5} \right)}}} \right):\frac{{2x – 5}}{{x\left( {x + 5} \right)}} + \frac{x}{{5 – x}}\\
    = \frac{{10x – 25}}{{x\left( {x – 5} \right)\left( {x + 5} \right)}}.\frac{{x\left( {x + 5} \right)}}{{2x – 5}} + \frac{x}{{5 – x}}\\
    = \frac{5}{{x – 5}} – \frac{x}{{x – 5}}\\
    = \frac{{5 – x}}{{x – 5}} =  – 1
   \end{array}\)

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2. ` A = x/(x^2 – 25) – (x-5)/(x^2 +5x) :  (2x-5)/(x^2+5x) + x/(5-x)`

   ` = x/((x-5)(x+5)) – (x-5)/(x(x+5)) : (2x-5)/(x(x+5)) + x/(5-x)`

   ` = (x^2)/(x(x-5)(x+5)) – ( (x-5)^2)/((x(x-5)(x+5)) : (2x-5)/(x(x+5)) + x/(5-x)`

   ` = (x^2-  x^2 + 10x – 25)/(x(x-5)(x+5)) : (2x-5)/(x(x+5)) + x/(5-x)`

   ` = ( 5(2x-5))/(x(x-5)(x+5)) . (x(x+5))/(2x-5) + x/(5-x)`

   ` = 5/(x-5) + x/(5-x)`

   ` = 5/(x-5)  -x/(x-5)`

   ` = (5-x)/(x-5)`

   `= -1`

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