a/ 3xy(x + y) – (x+y)(x^2 + y^2 + 2xy) + y^3 = 27 b/ (8x – 3)(3x + 2) – (4x + 7)(x + 4) = (2x + 1)(5x – 1) – 33

a/ 3xy(x + y) – (x+y)(x^2 + y^2 + 2xy) + y^3 = 27
b/ (8x – 3)(3x + 2) – (4x + 7)(x + 4) = (2x + 1)(5x – 1) – 33

0 bình luận về “a/ 3xy(x + y) – (x+y)(x^2 + y^2 + 2xy) + y^3 = 27 b/ (8x – 3)(3x + 2) – (4x + 7)(x + 4) = (2x + 1)(5x – 1) – 33”

  1. a)3xy(x + y) – (x+y)(x^2 + y^2 + 2xy) + y^3 = 27

    =3x²y+3xy²-x³+xy²+2x²y+x²y+y³+2xy²+y³=27

    =2y³+6xy²+6x²y-x³

    xin lỗi bạn câu b mình ko biết làm chúc bạn học tốt

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  2. `a,3xy(x + y) – (x+y)(x^2 + y^2 + 2xy) + y^3 = 27`

    `3x²y+3xy²-x³+xy²+2x²y+x²y+y³+2xy²+y³=27`

    `2y³+6xy²+6x²y-x³`

    `b,(8x−3)(3x+2)−(4x+7)(x+4)+(2x+1)(1−5x)=−33` 

    `⇔3x(8x−3)+2(8x−3)−[x(4x+7)+4(4x+7)]+(2x+1)−5x(2x+1)+33=0`

    `⇔24x^2−9x+16x−6−(4x^2+7x+16x+28)+2x+1−10x^2−5x+33=0`

    ⇔24x^2−9x+16x−6−4x^2−7x−16x−28+2x+1−10x^2−5x+33=0`

    `⇔10x^2−19x=0`

    `⇔x(10x−19)=0`

    `⇔`\(\left[ \begin{array}{l}x=0\\10x-19=0\end{array} \right.\)

    `⇔` \(\left[ \begin{array}{l}x=0\\x=\dfrac{19}{10}\end{array} \right.\)

    Xin hay nhất !

     

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