a,Tìm x:cx2 + 36 = 12x b, Phân tích đa thức thành nhân tử : x3 + 1 – x2 – x (x + y)3 – x3 – y3

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1. `a)x^2+36=12x`

   `⇔x^2-12x+36=0`

   `⇔x^2-2.x.6+6^2=0`

   `⇔(x-6)^2=0`

   `⇔x-6=0`

   `⇔x=6`

   Vậy `x=6`

   `b)x^3+1-x^2-x`

   `=(x^3-x^2)-(x-1)`

   `=x^2(x-1)-(x-1)`

   `=(x^2-1)(x-1)`

   `=(x+1)(x-1)(x-1)`

   `=(x+1)(x-1)^2`

   `c)(x+y)^3-x^3-y^3`

   `=x^3+3x^2y+3xy^2+y^3-x^3-y^3`

   `=(x^3-x^3)+(y^3-y^3)+3x^2y+3xy^2`

   `=3xy(x+y)`

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2. a, `x²+36=12x`

   `⇔x²-12x+36=0`

   `⇔(x-6)²=0`

   `⇔x-6=0`

   `⇒x=6`

   Vậy `x=6`

   b1, `x³+1-x²-x`

   `=(x+1)(x²-x+1)-x(x+1)`

   `=(x+1)(x²-x+1-x)`

   `=(x+1)(x²-2x+1)`

   `=(x+1)(x-1)²`

   b2, `(x+y)³ – x³-y³`

   `=x³+3x²y +3xy²+y³-x³-y³`

   `=(x³-x³)+(y³-y³)+3x²y+3xy²`

   `=3xy(x+y)`

   `#SADTODAY_`

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