Phân tích ĐTTNT (x^2-x+1)^2-5x(x^2-x+1)+4x^2

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Phân tích ĐTTNT
(x^2-x+1)^2-5x(x^2-x+1)+4x^2

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Camila 4 tháng 2021-08-11T11:11:05+00:00 2 Answers 5 views 0

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    0
    2021-08-11T11:12:49+00:00

     Ta có : $(x^2-x+1)^2-5x.(x^2-x+1)+4x^2$

    $= (x^2-x+1)^2-x(x^2+x+1)-4x.(x^2-x+1)+4x^2$

    $ = (x^2-x+1).(x^2-x+1-x)-4x.(x^2-x+1-x)$

    $ = (x^2-x+1-x).(x^2-x+1-4x)$

    $ = (x^2-2x+1).(x^2-5x+1)$

    $ = (x-1)^2.(x^2-5x+1)$

    0
    2021-08-11T11:12:57+00:00

    Đáp án: $(x-1)^2(x^2-5x+1)$

     

    Giải thích các bước giải:

    $(x^2-x+1)^2-5x(x^2-x+1)+4x^2$

    $=(x^2-x+1)^2-4x(x^2-x+1)-x(x^2-x+1)+4x^2$

    $=(x^2-x+1)(x^2-x+1-4x)-x(x^2-x+1-4x)$

    $=(x^2-x+1-x)(x^2-x+1-4x)$

    $=(x^2-2x+1)(x^2-5x+1)$

    $=(x-1)^2(x^2-5x+1)$

     

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