Question

[Phạm Quốc Tiến]

1/phân tích đa thức thành nhân tử 

4-32x³

2/thực hiện phép tính

a/(-x²+6x³-26x+21)/(2x-3)

b/\(\left(\frac{x}{xy-y^2}-\frac{2x-y}{xy-x^2}\right)\)/\( \left(\frac{1}{x}+\frac{1}{y}\right)\)

Answer 1

[Minh Hiền]

1. 4-32x³

= 4.(1-8x³)

= 4.[1³-(2x)³ ]

= 4.(1-2x).(1+2x+4x²)

2. b. \(\left(\frac{x}{xy-y^2}-\frac{2x-y}{xy-x^2}\right):\left(\frac{1}{x}+\frac{1}{y}\right)\)

\(=\left[\frac{x}{y\left(x-y\right)}+\frac{2x-y}{x\left(x-y\right)}\right]:\left(\frac{y}{xy}+\frac{x}{xy}\right)\)

\(=\left[\frac{x.x}{y\left(x-y\right).x}+\frac{\left(2x-y\right).y}{x\left(x-y\right).y}\right]:\left(\frac{x+y}{xy}\right)\)

\(=\left[\frac{x^2+2xy-y^2}{xy\left(x-y\right)}\right]:\left(\frac{x+y}{xy}\right)\)

\(=\left[\frac{-\left(x-y\right)^2}{xy\left(x-y\right)}\right].\frac{xy}{x+y}\)

\(=\frac{-\left(x-y\right)}{xy}.\frac{xy}{x+y}\)

\(=\frac{y-x}{x+y}\)