Question

a) (\(8x^3\) - \(\dfrac{1}{125}y^3\)) : (\(4x^2+\dfrac{1}{25}y^2+\dfrac{2}{5}xy\))

b) (\(x^3-6x^2+2x+15\)) : (\(x-5\))

Answer 1

Hằng đẳng thức mà tương ạ! :v

a, \(\dfrac{8x^3-\dfrac{1}{125}y^3}{4x^2+\dfrac{1}{25}y^2+\dfrac{2}{5}xy}\)

\(=\dfrac{\left(2x-\dfrac{1}{5}y\right)\left(4x^2+\dfrac{2}{5}xy+\dfrac{1}{25}y^2\right)}{4x^2+\dfrac{1}{25}y^2+\dfrac{2}{5}xy}=2x-\dfrac{1}{5}y\)

b, \(\dfrac{x^3-6x^2+2x+15}{x-5}\)

\(=\dfrac{x^3-5x^2-x^2+5x-3x+15}{x-5}\)

\(=\dfrac{x^2\left(x-5\right)-x\left(x-5\right)-3\left(x-5\right)}{x-5}\)

\(=\dfrac{\left(x-5\right)\left(x^2-x-3\right)}{\left(x-5\right)}=x^2-x-3\)

Rồi ạ :v!

Answer 2

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Answer 3

đề bài là rút gọn biểu thức à??????

Answer 4

à bạn kia làm rồi hihi nãy chưa nhìn thấy

Answer 5

a)

\(\left(8x^3-\dfrac{1}{125}y^3\right):\left(4x^2+\dfrac{1}{25}y^2+\dfrac{2}{5}xy\right)\)

=\(\left(2x-\dfrac{1}{5}y\right)\left(4x^2+\dfrac{2}{5}xy+\dfrac{1}{25}y^2\right):\left(4x^2+\dfrac{2}{5}xy+\dfrac{1}{25}y^2\right)\)

=\(2x-\dfrac{1}{5}y\)

b)

\(x^3-6x^2+2x+15:\left(x-5\right)\)

=\(x^3-5x^2-x^2+5x-3x+15:\left(x-5\right)\)

=\(x^2\left(x-5\right)-x\left(x-5\right)-3\left(x-5\right)\)

=\(\left(x-5\right)\left(x^2-x+3\right):\left(x-5\right)=x^2-x+3\)

Answer 6

Anh hiếu làm ròi em ko làm nữa :D