Giải gấp hộ mình vs ạ mơn trc

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Giải gấp hộ mình vs ạ mơn trc

giai-gap-ho-minh-vs-a-mon-trc

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Bơ 5 years 2020-11-04T23:35:14+00:00 1 Answers 59 views 0

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  1. Sigrid
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    2020-11-04T23:37:12+00:00
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    Giải thích các bước giải:

    Ta có:

    \(\begin{array}{l}
    1,\\
    a,\\
    3{x^2} – 6x = 3x.\left( {x – 2} \right)\\
    b,\\
    20{x^2}y – 12xy = 4xy\left( {5x – 3} \right)\\
    c,\\
    3x\left( {x – 6} \right) – 2\left( {x – 6} \right) = \left( {x – 6} \right)\left( {3x – 2} \right)\\
    d,\\
    7x\left( {x – y} \right) – \left( {y – x} \right) = 7x\left( {x – y} \right) – y + x = 7x\left( {x – y} \right) + \left( {x – y} \right) = \left( {x – y} \right)\left( {7x + 1} \right)\\
    e,\\
    25{x^2} – 4 = {\left( {5x} \right)^2} – {2^2} = \left( {5x – 2} \right)\left( {5x + 2} \right)\\
    g,\\
    16 – {\left( {x – y} \right)^2} = {4^2} – {\left( {x – y} \right)^2} = \left[ {4 – \left( {x – y} \right)} \right].\left[ {4 + \left( {x – y} \right)} \right] = \left( { – x + y + 4} \right)\left( {x – y + 4} \right)\\
    h,\\
    2x + 2y – {x^2} – xy = \left( {2x + 2y} \right) – \left( {{x^2} + xy} \right) = 2\left( {x + y} \right) – x\left( {x + y} \right) = \left( {x + y} \right)\left( {2 – x} \right)\\
    f,\\
    {x^2} + 2xy + {y^2} – xz – yz = \left( {{x^2} + 2xy + {y^2}} \right) – \left( {xz + yz} \right) = {\left( {x + y} \right)^2} – z\left( {x + y} \right) = \left( {x + y} \right)\left( {x + y – z} \right)\\
    j,\\
    {x^2}y – {x^3} – 9y + 9x = \left( {{x^2}y – {x^3}} \right) – \left( {9y – 9x} \right) = {x^2}\left( {y – x} \right) – 9\left( {y – x} \right) = \left( {y – x} \right).\left( {{x^2} – 9} \right) = \left( {y – x} \right).\left( {x – 3} \right)\left( {x + 3} \right)\\
    2,\\
    a,\\
    5{x^3} + 10{x^2} = 0\\
     \Leftrightarrow 5{x^2}\left( {x + 2} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    {x^2} = 0\\
    x + 2 = 0
    \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
    x = 0\\
    x =  – 2
    \end{array} \right.\\
    b,\\
    3x\left( {x – 1} \right) – x + 1 = 0\\
     \Leftrightarrow 3x\left( {x – 1} \right) – \left( {x – 1} \right) = 0\\
     \Leftrightarrow \left( {x – 1} \right)\left( {3x – 1} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    x – 1 = 0\\
    3x – 1 = 0
    \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
    x = 1\\
    x = \frac{1}{3}
    \end{array} \right.\\
    c,\\
    {x^3} – {x^2} – x + 1 = 0\\
     \Leftrightarrow \left( {{x^3} – {x^2}} \right) – \left( {x – 1} \right) = 0\\
     \Leftrightarrow {x^2}\left( {x – 1} \right) – \left( {x – 1} \right) = 0\\
     \Leftrightarrow \left( {x – 1} \right)\left( {{x^2} – 1} \right) = 0\\
     \Leftrightarrow \left( {x – 1} \right)\left( {x – 1} \right)\left( {x + 1} \right) = 0\\
     \Leftrightarrow {\left( {x – 1} \right)^2}\left( {x + 1} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    x – 1 = 0\\
    x + 1 = 0
    \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
    x = 1\\
    x =  – 1
    \end{array} \right.
    \end{array}\)

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