a) 12x 2x
______ + _____
x^2-9 x+3
b) 1-x x-2
______ + _______
x-3 x-3
a) 12x 2x ______ + _____ x^2-9 x+3 b) 1-x x-2 ______ + _______ x-3 x-3
By Caroline
By Caroline
a) 12x 2x
______ + _____
x^2-9 x+3
b) 1-x x-2
______ + _______
x-3 x-3
`(12x)/(x^2-9)+(2x)/(x+3)` Điều kiện xác định: `x\ne±3`
`=(12x)/((x-3)(x+3))+\frac{2x(x-3)}{(x-3)(x+)}`
`=\frac{12x+2x^2-6x}{(x-3)(x+3)}`
`=\frac{2x^2+6x}{(x-3)(x+3)}`
`=\frac{2x(x+3)}{(x-3)(x+3)}`
`=\frac{2x}{x-30}`
`b)` `\frac{1-x}{x-3}+\frac{x-2}{x-3}` Điều kiện xác định: `n\ne3`
`=\frac{1-x+x-2}{x-3}`
`=\frac{-1}{x-3}`
` a) ` ` \frac{12x}{x^2 – 9} + \frac{2x}{x + 3} `
` = \frac{12x + 2x(x – 3)}{(x – 3)(x + 3)} `
` = \frac{12x + 2x^2 – 6x}{(x – 3)(x + 3)} `
` = \frac{2x^2 – 6x}{(x – 3)(x + 3)} `
` = \frac{2x(x – 3)}{(x – 3)(x + 3)} `
` = \frac{2x}{x + 3} `
` b) ` ` \frac{1 – x}{x – 3} + \frac{x – 2}{x – 3} `
` = \frac{1 – x + x – 2}{x – 3} `
` = \frac{-1}{x – 3} `