Giúp mk vs ạ cảm ơn bạn nhiều lắm

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Giúp mk vs ạ cảm ơn bạn nhiều lắm

giup-mk-vs-a-cam-on-ban-nhieu-lam

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Tryphena 9 months 2020-11-01T09:09:18+00:00 1 Answers 63 views 0

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    2020-11-01T09:11:01+00:00

    Đáp án:

    $\begin{array}{l}
    a){x^2} + 5x – 6\\
     = {x^2} + 6x – x – 6\\
     = \left( {x + 6} \right)\left( {x – 1} \right)\\
    b){x^2} – 7x + 10\\
     = {x^2} – 2x – 5x + 10\\
     = \left( {x – 2} \right)\left( {x – 5} \right)\\
    c){x^3} + {x^2} – 3x – 27\\
     = {x^3} – 27 + {x^2} – 3x\\
     = \left( {x – 3} \right)\left( {{x^2} + 3x + 9} \right) + x\left( {x – 3} \right)\\
     = \left( {x – 3} \right)\left( {{x^2} + 3x + 9 + x} \right)\\
     = \left( {x – 3} \right)\left( {{x^2} + 4x + 9} \right)\\
    d)2{x^3} + 3{x^2} – 2x\\
     = x.\left( {2{x^2} + 3x – 2} \right)\\
     = x\left( {2{x^2} + 4x – x – 2} \right)\\
     = x\left( {x + 2} \right)\left( {2x – 1} \right)\\
    e){x^2} – xz – 9{y^2} + 3yz\\
     = {x^2} – 9{y^2} + 3yz – xz\\
     = \left( {x – 3y} \right)\left( {x + 3y} \right) – z\left( {x – 3y} \right)\\
     = \left( {x – 3y} \right)\left( {x + 3y – z} \right)\\
    f)\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right)\left( {x + 4} \right) – 120\\
     = \left( {x + 1} \right)\left( {x + 4} \right)\left( {x + 2} \right)\left( {x + 3} \right) – 120\\
     = \left( {{x^2} + 5x + 4} \right)\left( {{x^2} + 5x + 6} \right) – 120\\
    Dat:{x^2} + 5x + 4 = a\\
     \Rightarrow a.\left( {a + 2} \right) – 120\\
     = {a^2} + 2a – 120\\
     = {a^2} + 12a – 10a – 120\\
     = \left( {a + 12} \right)\left( {a – 10} \right)\\
     = \left( {{x^2} + 5x + 4 + 12} \right)\left( {{x^2} + 5x + 4 – 10} \right)\\
     = \left( {{x^2} + 5x + 16} \right)\left( {{x^2} + 5x – 6} \right)\\
    g){x^3} – 5{x^2} – 4x + 20\\
     = {x^2}\left( {x – 5} \right) – 4\left( {x – 5} \right)\\
     = \left( {x – 5} \right)\left( {{x^2} – 4} \right)\\
     = \left( {x – 5} \right)\left( {x – 2} \right)\left( {x + 2} \right)
    \end{array}$

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