Question

Đã trả lời bởi chuyên gia

Answer 1

`c,`

`(x^2-xy)/(y^2-x^2)`

`= (x(x-y))/((y-x)(y+x)`

`= -(x(x-y))/((x-y)(y+x)`

`= -x/(y+x)`

`d,`

`(x^2-9)/(3x-x^2)`

`= ((x-3)(x+3))/(x(3-x)`

`= -((x-3)(x+3))/(x(x-3)`

`= -(x+3)/x`

`e,` Giống ý `c`

`f,`

`(2x-4y)/(x^2-4y^2)`

`= (2(x-2y))/((x-2y)(x+2y))`

`= 2/(x+2y)`

`g)`
`(3xy+3x)/(9y+9) = [3.(y+1)]/[9.(y+1)] = 3/9 = 1/3`
`h)`
`(2x^2 + 2x)/(x+1) = [2x.(x+1)]/(x+1) = 2x`
`i)`
`(x^2 - xy )/(3x^2 - 3xy) = [x.(x-y)]/[3x.(x-y)] = x/(3x) = 1/3`
`j)`
`(x^2 - xy )/(5y^2 - 5xy) = [x.(x-y)]/[5y.(y-x)] = -[x.(y-x)]/[5y.(y-x)] = -x/(5y)`
`k)`
`(10xy - 5x^2 )/(2x^2 - 8y^2 ) = [5x.(2y-x)]/[2.(x^2 - 4y^2 )] = [5x.(2y-x)]/[2.(x-2y).(x+2y)] = -[5x.(x-2y)]/[2.(x-2y).(x+2y)] = -(5x)/[2.(x+2y)]`

Answer 2

c) $\frac{x^{2} - x y}{y^{2} - x^{2}} = \frac{x (x - y)}{- (x^{2} - y^{2})} = - \frac{x (x - y)}{(x - y) (x + y)} = - \frac{x}{x + y}$

d) $\frac{x^{2} - 9}{3 x - x^{2}} = \frac{x^{2} - 3^{2}}{- (x^{2} - 3 x)} = - \frac{(x - 3) (x + 3)}{x (x - 3)} = - \frac{x + 3}{x}$

e) Giống câu c nhé

f) $\frac{2 x - 4 y}{x^{2} - 4 y^{2}} = \frac{2 (x - 2 y)}{x^{2} - {(2 y)}^{2}} = \frac{2 (x - 2 y)}{(x - 2 y) (x + 2 y)} = \frac{2}{x + 2 y}$

g) $\frac{3 x y + 3 x}{9 y + 9} = \frac{3 x (y + 1)}{9 (y + 1)} = \frac{3 x}{9} = \frac{x}{3}$

h) $\frac{2 x^{2} + 2 x}{x + 1} = \frac{2 x (x + 1)}{x + 1} = 2 x$

i) $\frac{x^{2} - x y}{3 x^{2} - 3 x y} = \frac{x (x - y)}{3 x (x - y)} = \frac{x}{3 x} = \frac{1}{3}$

j) $\frac{x^{2} - x y}{5 y^{2} - 5 x y} = \frac{x (x - y)}{- 5 y (x - y)} = - \frac{x}{5 y}$

k) $\frac{10 x y - 5 x^{2}}{2 x^{2} - 8 y^{2}} = \frac{- (5 x^{2} - 10 x y)}{2 (x^{2} - 4 y^{2})} = - \frac{5 x (x - 2 y)}{2 (x - 2 y) (x + 2 y)} = - \frac{5 x}{2 (x + 2 y)} = - \frac{5 x}{2 x + 4 y}$

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Answer 3