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

Toán Lớp 9: A=((√x – 1)/(√x + 1) – (√x + 1)/(√x – 1)):(1/2√x – √x/2). Rút gọn A

Toán Lớp 9: A=((√x – 1)/(√x + 1) – (√x + 1)/(√x – 1)):(1/2√x – √x/2). Rút gọn A

Comments ( 2 )

  1. = \frac{(\sqrt{x}-1)^2-(\sqrt{x}+1)^2){(\sqrt{x}+1)(\sqrt{x}-1)}:(\frac{1}{2\sqrt{x}}-\frac{\sqrt{x}}{2})
    = \frac{-2.2\sqrt{x}}{(\sqrt{x}+1)(\sqrt{x}-1)}:(\frac{1}{2\sqrt{x}}-\frac{\sqrt{x}}{2})
    =- \frac{4\sqrt{x}}{(\sqrt{x}+1)(\sqrt{x}-1)}:\frac{1-x}{2\sqrt{x}}
    =- \frac{4\sqrt{x}}{(\sqrt{x}+1)(\sqrt{x}-1)}.\frac{2\sqrt{x}}{1-x}
    = -\frac{8x}{(x-1)[-(x-1)]}
    = -\frac{8x}{-(x-1)^2}
    = \frac{8x}{(x-1)^2}

  2. Giải đáp:
    \(\dfrac{{8x}}{{{{\left( {x – 1} \right)}^2}}}\)
    Lời giải và giải thích chi tiết:
    \(\begin{array}{l}
    DK:x > 0;x \ne 1\\
    A = \left( {\dfrac{{\sqrt x  – 1}}{{\sqrt x  + 1}} – \dfrac{{\sqrt x  + 1}}{{\sqrt x  – 1}}} \right):\left( {\dfrac{1}{{2\sqrt x }} – \dfrac{{\sqrt x }}{2}} \right)\\
     = \dfrac{{{{\left( {\sqrt x  – 1} \right)}^2} – {{\left( {\sqrt x  + 1} \right)}^2}}}{{\left( {\sqrt x  – 1} \right)\left( {\sqrt x  + 1} \right)}}:\dfrac{{1 – x}}{{2\sqrt x }}\\
     = \dfrac{{x – 2\sqrt x  + 1 – x – 2\sqrt x  – 1}}{{\left( {\sqrt x  – 1} \right)\left( {\sqrt x  + 1} \right)}}.\dfrac{{2\sqrt x }}{{1 – x}}\\
     = \dfrac{{ – 4\sqrt x }}{{\left( {x – 1} \right)}}.\dfrac{{2\sqrt x }}{{1 – x}}\\
     = \dfrac{{8x}}{{{{\left( {x – 1} \right)}^2}}}
    \end{array}\)

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

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