Đáp án: `a) x+1`
`b)`` (2)/(x+1)`
Giải thích các bước giải:
`a)(12x ³ - 28x ² + 21x - 5) : (6x - 5) - (2x ² - 4x)`
`=(12x ³ - 18x ²-10x^2 + 6x + 15x - 5) : (6x - 5) - (2x ² - 4x)`
`=[(12x ³ - 18x ²+ 6x)+(-10x^2 + 15x - 5)] : (6x - 5) - (2x ² - 4x)`
`=[6x.(2x^2-3x+1)-5.(2x^2-3x+1) ] : (6x - 5) - (2x ² - 4x)`
`=[(6x-5).(2x^2-3x+1)] : (6x - 5) - (2x ² - 4x)`
`=(2x^2-3x+1) - (2x ² - 4x)=2x^2-3x+1-2x ² + 4x=x+1`
.
`b)``((x+1)/(x-3)+(5x-39)/(x^2-9)-(11)/(x+3)):(x^2+2x+1)/(2x+6)`
`=(((x+1)(x+3))/((x-3)(x+3))+(5x-39)/((x-3)(x+3))-(11.(x-3))/((x-3)(x+3))):(x+1)^2/(2(x+3))`
`=((x+1)(x+3)+5x-39-11.(x-3))/((x-3)(x+3)):(x+1)^2/(2(x+3))`
`=(x^2+4x+3+5x-39-11x+33)/((x-3)(x+3)):(x+1)^2/(2(x+3))`
`=(x^2-2x-3)/((x-3)(x+3)):(x+1)^2/(2(x+3))=((x-3)(x+1))/((x-3)(x+3)).(2(x+3))/(x+1)^2`
`=(x+1)/(x+3).(2(x+3))/((x+1)(x+1))=(2)/(x+1)`
