39 a) $frac{1}{9x-18}$ + $frac{22-7x}{72-18x^2}$ + $frac{5}{12x+24}$ b)$frac{x+1}{x-1}$ – $frac{x-1}{x+1}$ – $frac{4}{1-x^2}$ mới nhất

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    2021-12-07T19:24:20+00:00 07/12/2021 at 19:24
    Reply

    Đáp án:

    $begin{array}{l}
    a)Dkxd:x ne 2;x ne  – 2\
    dfrac{1}{{9x – 18}} + dfrac{{22 – 7x}}{{72 – 18{x^2}}} + dfrac{5}{{12x + 24}}\
     = dfrac{1}{{9left( {x – 2} right)}} + dfrac{{22 – 7x}}{{18.left( {4 – {x^2}} right)}} + dfrac{5}{{12.left( {x + 2} right)}}\
     = dfrac{{4.left( {x + 2} right) + left( {7x – 22} right).2 + 3.left( {x – 2} right).5}}{{36left( {x – 2} right)left( {x + 2} right)}}\
     = dfrac{{4x + 8 + 14x – 44 + 15x – 30}}{{36left( {x – 2} right)left( {x + 2} right)}}\
     = dfrac{{33x – 66}}{{36left( {x – 2} right)left( {x + 2} right)}}\
     = dfrac{{33.left( {x – 2} right)}}{{36left( {x – 2} right)left( {x + 2} right)}}\
     = dfrac{{11}}{{12left( {x + 2} right)}}\
    b)Dkxd:x ne 1;x ne  – 1\
    dfrac{{x + 1}}{{x – 1}} – dfrac{{x – 1}}{{x + 1}} – dfrac{4}{{1 – {x^2}}}\
     = dfrac{{{{left( {x + 1} right)}^2} – {{left( {x – 1} right)}^2} + 4}}{{left( {x – 1} right)left( {x + 1} right)}}\
     = dfrac{{{x^2} + 2x + 1 – {x^2} + 2x – 1 + 4}}{{left( {x + 1} right)left( {x – 1} right)}}\
     = dfrac{{4x + 4}}{{left( {x + 1} right)left( {x – 1} right)}}\
     = dfrac{{4left( {x + 1} right)}}{{left( {x + 1} right)left( {x – 1} right)}}\
     = dfrac{4}{{x – 1}}
    end{array}$