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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: ???? = ????−1/2 :( ????2+2/????3−1 + ????/????2+????+1 + 1/1−???? )

Toán Lớp 8: ???? = ????−1/2 🙁 ????2+2/????3−1 + ????/????2+????+1 + 1/1−???? )

Comments ( 1 )

  1. Giải đáp:
    \(\dfrac{{{x^2} + x + 1}}{2}\)
    Lời giải và giải thích chi tiết:
    \(\begin{array}{l}
    DK:x \ne 1\\
    M = \dfrac{{x – 1}}{2}:\left( {\dfrac{{{x^2} + 2}}{{{x^3} – 1}} + \dfrac{x}{{{x^2} + x + 1}} + \dfrac{1}{{1 – x}}} \right)\\
     = \dfrac{{x – 1}}{2}:\dfrac{{{x^2} + 2 + x\left( {x – 1} \right) – \left( {{x^2} + x + 1} \right)}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}\\
     = \dfrac{{x – 1}}{2}.\dfrac{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}{{{x^2} + 2 + {x^2} – x – {x^2} – x – 1}}\\
     = \dfrac{{x – 1}}{2}.\dfrac{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}{{{x^2} – 2x + 1}}\\
     = \dfrac{{x – 1}}{2}.\dfrac{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}{{{{\left( {x – 1} \right)}^2}}}\\
     = \dfrac{{{x^2} + x + 1}}{2}
    \end{array}\)

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222-9+11+12:2*14+14 = ? ( )

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