Đáp án ` + ` Giải thích các bước giải:

`#\text{Day 3}`

` a) `

` (2x + 1/3)(-1/5x - 2/3) = 0 `

` -> [(2x + 1/3 = 0),(-1/5x - 2/3 = 0):} `

` -> [(2x = -1/3),(-1/5x = 2/3):} `

` -> [(x = -1/6),(x = -10/3):} `

Vậy ` S = {-1/6 ; -10/3} `

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` b) `

` x(3x + 5) - 6x - 10 = 0 `

` -> x(3x + 5) - 2(3x + 5) = 0 `

` -> (3x + 5)(x - 2) = 0 `

` -> [(3x + 5 = 0),(x - 2 = 0):} `

` -> [(3x = -5),(x = 2):} `

` -> [(x = -5/3),(x = 2):} `

Vậy ` S = {-5/3 ; 2} `

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` c) `

` (2x - 5)^2 - 9x^2 = 0 `

` -> (2x - 5)^2 - (3x)^2 = 0 `

` -> [(2x - 5) - 3x] * [(2x - 5) + 3x] = 0 `

` -> (-x - 5)(5x - 5) = 0 `

` -> [(-x - 5 = 0),(5x - 5 = 0):} `

` -> [(x = -5),(5x = 5):} `

` -> [(x = -5),(x = 1):} `

Vậy ` S = {-5 ; 1} `

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` d) `

` (3x)/(2x + 1) - (4x^2 + 1)/(2x^2 - 1) = x/(2x - 1) `

ĐKXĐ: ` x \ne ±1/2 `

` -> 3x(2x - 1) - (4x^2 + 1) = x(2x + 1) `

` -> 6x^2 - 3x - 4x^2 - 1 = 2x^2 + x `

` -> 2x^2 - 3x - 1 = 2x^2 + x `

` -> -3x - 1 = x `

` -> -4x = 1 `

` -> x = -1/4 `

Vậy ` S = {-1/4} `

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` e) `

` (x + 3)/(x - 3) - (x^2)/(x^2 - 9) = (-6)/(x + 3) `

ĐKXĐ: ` x \ne ±3 `

` -> (x + 3)^2 - x^2 = -6(x - 3) `

` -> x^2 + 6x + 9 - x^2 = -6x + 18 `

` -> 6x + 9 = -6x + 18 `

` -> 12x = 9 `

` -> x = 3/4 `

Vậy ` S = {3/4} `

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` g) `

` 1/(x + 1) - x/(x^2 - x + 1) = (3x)/(x^3 + 1) `

ĐKXĐ: ` x \ne -1 `

` -> (x^2 - x + 1)/(x^3 + 1) - (x(x + 1))/(x^3 + 1) = (3x)/(x^3 + 1) `

` -> x^2 - x + 1 - x(x + 1) = 3x `

` -> x^2 - x + 1 - x^2 - x = 3x `

` -> -2x + 1 = 3x `

` -> 5x = 1 `

` -> x = 1/5 `

Vậy ` S = {1/5} `