1. Tìm x biết a, $(y+3)^3 – (y+1)^3 = 56$ b, $(y-2)^3 – (y-3)(y^2 + 3y + 9) + 6(y+1)^2 = 49$ c, $(x-1)^3 + 1 + 3x(x-4) = 0$ d, $12x – x^2 – 36 = 0$ Là

1. Tìm x biết
a, $(y+3)^3 – (y+1)^3 = 56$
b, $(y-2)^3 – (y-3)(y^2 + 3y + 9) + 6(y+1)^2 = 49$
c, $(x-1)^3 + 1 + 3x(x-4) = 0$
d, $12x – x^2 – 36 = 0$
Làm hết nhé – :))

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  1. Đáp án:

     

    Giải thích các bước giải:

    `a,(y+3)^3-(y+1)^3=56`

    `<=>[(y+3)-(y+1)][(y+3)^2+(y+3)(y+1)+(y+1)^2]-56=0`

    `<=>2(y^2+6y+9+y^2+4y+3+y^2+2y+1)-56=0`

    `<=>y^2+6y+9+y^2+4y+3+y^2+2y+1-28=0`

    `<=>3y^2+12y-15=0`

    `<=>y^2+4y-5=0`

    `<=>(y-1)(y+5)=0`

    `<=>`\(\left[ \begin{array}{l}y=1\\y=-5\end{array} \right.\)

    `b,(y-2)^3-(y-3)(y^2+3y+9)+6(y+1)^2=49`

    `<=>y^3-6y^2+12y-8-y^3+27+6y^2+12y+6-49=0`

    `<=>24y-24=0`

    `<=>y-1=0`

    `<=>y=1`

    `c,(x-1)^3+1+3x(x-4)=0`

    `<=>(x-1+1)[(x-1)^2-(x-1)+1]+3x(x-4)=0`

    `<=>x(x^2-2x+1-x+1+1]+3x(x-4)=0`

    `<=>x(x^2-2x+1-x+1+1+3x-12)=0`

    `<=>x(x^2-9)=0`

    `<=>`\(\left[ \begin{array}{l}x=0\\x^2-9=0\end{array} \right.\)

    `<=>`\(\left[ \begin{array}{l}x=0\\\left[ \begin{array}{l}x=3\\x=-3\end{array} \right.\end{array} \right.\)

    `d,12x-x^2-36=0`

    `<=>x^2-12x+36=0`

    `<=>(x-6)^2=0`

    `<=>x-6=0`

    `<=>x=6`

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