Hướng dẫn trả lời:
Bài 1:
19) `12x^2ycdot(x - 5) + 20xy^2cdot(5 - x)`
`= 12x^2ycdot(x - 5) + 20xy^2cdot[- (x - 5)]`
`= 12x^2ycdot(x - 5) - 20xy^2cdot(x - 5)`
`= (12x^2y - 20xy^2)cdot(x - 5)`
`= 4xycdot(3x - 5y)cdot(x - 5)`
20) `12ycdot(2x - 5) + 6xycdot(5 - 2x)`
`= 12ycdot(2x - 5) + 6xycdot[-(2x - 5)]`
`= 12ycdot(2x - 5) - 6xycdot(2x - 5)`
`= (12y - 6xy)cdot(2x - 5)`
`= 6ycdot(2 - x)cdot(2x - 5)`
Bài 2:
a) `2x^3 - 50x = 0`
`↔ 2xcdot(x^2 - 25) = 0`
`↔ 2xcdot(x^2 - 5^2) = 0`
`↔ 2xcdot(x + 5)cdot(x - 5) = 0`
`↔ [(2x = 0),(x + 5 = 0),(x - 5 = 0):}`
`↔ [(x = 0),(x = - 5),(x = 5):}`
Vậy `x = 0` hoặc `x = - 5` hoặc `x = 5`
b) `2xcdot(3x - 5) - (5 - 3x) = 0`
`↔ 2xcdot(3x - 5) - [- (3x - 5)] = 0`
`↔ 2xcdot(3x - 5) + 1cdot(3x - 5) = 0`
`↔ (2x + 1)cdot(3x - 5) = 0`
`↔ [(2x + 1 = 0),(3x - 5 = 0):}`
`↔ [(2x = - 1),(3x = 5):}`
`↔ [(x = - 1/2),(x = 5/3):}`
Vậy `x = -1/2` hoặc `x = 5/3`
c) `x^3 - 9x = 0`
`↔ xcdot(x^2 - 9) = 0`
`↔ xcdot(x^2 - 3^2) = 0`
`↔ xcdot(x + 3)cdot(x - 3) = 0`
`↔ [(x = 0),(x + 3 = 0),(x - 3 = 0):}`
`↔ [(x = 0),(x = - 3),(x = 3):}`
Vậy `x = 0` hoặc `x = -3` hoặc `x = 3`
d) `(2x - 1)^2 - 25 = 0`
`↔ (2x - 1) - 5^2 = 0`
`↔ [(2x - 1) + 5]cdot[(2x - 1) - 5] = 0`
`↔ (2x - 1 + 5)cdot(2x - 1 - 5) = 0`
`↔ (2x + 4)cdot(2x - 6) = 0`
`↔ 2cdot(x + 2)cdot2cdot(x - 3) = 0`
`↔ 4cdot(x + 2)cdot(x - 3) = 0`
`↔ (x + 2)cdot(x - 3) = 0`
`↔ [(x + 2 = 0),(x - 3 = 0):}`
`↔ [(x = - 2),(x = 3):}`
Vậy `x = -2` hoặc `x = 3`
e) `25x^2 - 2 = 0`
`↔ (5x)^2 - (sqrt{2})^2 = 0`
`↔ (5x + sqrt{2})cdot(5x - sqrt{2}) = 0`
`↔ [(5x + sqrt{2} = 0),(5x - sqrt{2} = 0):}`
`↔ [(5x = - sqrt{2}),(5x = sqrt{2}):}`
`↔ [(x = - sqrt{2}/5),(x = sqrt{2}/5):}`
Vậy `x = - sqrt{2}/5` hoặc `x = sqrt{2}/5`
f) `9cdot(3x - 2) = xcdot(2 - 3x)`
`↔ 9cdot(3x - 2) - xcdot(2 - 3x) = 0`
`↔ 9cdot(3x - 2) - xcdot[- (3x - 2)] = 0`
`↔ 9cdot(3x - 2) + xcdot(3x - 2) = 0`
`↔ (9 + x)cdot(3x - 2) = 0`
`↔ [(9 + x = 0),(3x - 2 = 0):}`
`↔ [(x = -9),(3x = 2):}`
`↔ [(x = -9),(x = 2/3):}`
Vậy `x = -9` hoặc `x = 2/3`
g) `3xcdot(x - 7) - 2cdot(x - 7) = 0`
`↔ (3x - 2)cdot(x - 7) = 0`
`↔ [(3x - 2 = 0),(x - 7 = 0):}`
`↔ [(3x = 2),(x = 7):}`
`↔ [(x = 2/3),(x = 7):}`
Vậy `x = 2/3` hoặc `x = 7`
h) `2x + 1 + xcdot(2x + 1) = 0`
`↔ 1cdot(2x + 1) + xcdot(2x + 1) = 0`
`↔ (1 + x)cdot(2x + 1) = 0`
`↔ [(1 + x = 0),(2x + 1 = 0):}`
`↔ [(x = -1),(2x = - 1):}`
`↔ [(x = -1),(x = - 1/2):}`
Vậy `x = -1` hoặc `x = -1/2`
