Đáp án+Giải thích các bước giải:
`c) (5x-1)/(3x+2) - (5x-7)/(3x-1)`
`= [(5x-1)(3x-1) - (5x-7)(3x+2)]/[(3x+2)(3x-1)]`
`= (15x² - 8x + 1 - 15x² + 11x + 14)/[(3x+2)(3x-1)]`
`= (3x + 15)/[(3x+2)(3x-1)]`
`= [3(x+5)]/[(3x+2)(3x-1)]`
`d) (2x+3)/(2x-3) - 3/(x-6) - 2/5`
`= [5(x-6)(2x+3) - 15(2x-3) - 2(x-6)(2x-3)]/[5(x-6)(2x-3)]`
`= (10x² + 15x - 60x - 90 - 30x + 45 - 4x² + 6x + 24x - 36) / [5(x-6)(2x-3)]`
`= (6x² - 45x - 81) / [5(x-6)(2x-3)]`
`= (3(2x² - 15x - 27)) / (5(x-6)(2x-3))`
`= (3(2x² + 3x - 18x - 27)) / (5(x-6)(2x-3))`
`= (3(x-9)(2x+3))/(5(x-6)(2x-3))`
`e) (x+1)/(x-2) - 5/(x+2) - 12/(4x) + 1`
`= (4x(x+1)(x+2) - 20x(x-2) - 12(x²-4) + 4x(x²-4))/(4x(x-2)(x+2))`
`= (4x³ + 12x² + 8x - 20x² + 40x - 12x² + 48 + 4x³ - 16x)/(4x(x-2)(x+2))`
`= (8x³ - 20x² +32x +48)/(4x(x-2)(x+2))`
`= (4(2x³ - 5x² + 8x + 12))/(4x(x-2)(x+2))`
`= (2x³ - 5x² + 8x + 12))/(x(x²-4))`
`f) 1/(x-1) + (2x²-5)/(x³-1) = 4/(x²+x+1)`
`ĐKXĐ : x` $\neq$ `1`
`-> (x² + x + 1 + 2x² - 5)/(x³-1) = (4(x-1))/(x³-1)`
`<=> 3x² + x - 4 = 4x - 4`
`<=> 3x² + x - 4x = 4-4`
`<=> 3x² - 3x = 0`
`<=> 3x(x-1) = 0`
`<=>` \(\left[ \begin{array}{l}x=0\\x=1(loại)\end{array} \right.\)
Vậy `x=0`
