Đáp án + Giải thích các bước giải:

`a) (x^2 - 5x)/(x - 5) = 5(x ne 5)` $\\$ `<=> (x^2 - 5x)/(x - 5) = [5(x - 5)]/(x - 5) => x^2 - 5x = 5x - 25` $\\$ `<=> x^2 - 5x - 5x + 25 = 0 <=> x^2 - 10x + 25 = 0` $\\$ `<=> (x - 5)^2 = 0 <=> x = 5(ktmđk)`

Vậy `S = emptyset`

`b) 1/(x - 1) - (3x^2)/(x^3 - 1) = (2x)/(x^2 + x + 1)(x ne 1)` $\\$ `<=> (x^2 + x + 1)/[(x - 1)(x^2 + x + 1)] - (3x^2)/[(x - 1)(x^2 + x + 1)] = (2x(x-1))/[(x - 1)(x^2 + x + 1)] ` $\\$ `=> x^2 + x + 1 - 3x^2 = 2x(x - 1)` $\\$ `<=> x^2 + x + 1 - 3x^2 = 2x^2 - 2x` $\\$ `<=> x^2 + x + 1 - 3x^2 - 2x^2 + 2x = 0` $\\$ `<=>-4x^2+3x+1=0` $\\$ `<=>-4x^2 + 4x - x + 1 = 0` $\\$ `<=> -4x(x - 1) - (x - 1) = 0` $\\$ `<=> (x - 1)(-4x - 1) = 0 <=> `\(\left[ \begin{array}{l}x=1(ktmđk)\\x=\frac{-1}{4}(tmđk)\end{array} \right.\)

Vậy `S = {-1/4}`

`c) 1/x + 2 = [(1 + 2)(x^2 + 1)]/x (x ne 0) <=> 1/x + (2x)/x = [(1 + 2)(x^2 + 1)]/x` $\\$ `<=> 1 +2x = (1 + 2)(x^2 + 1) <=> 1 + 2x = 3(x^2 + 1)` $\\$ `<=> 1 +2x = 3x^2 + 3 <=> 1 + 2x - 3x^2 - 3 = 0` $\\$ `<=> -3x^2 + 2x - 2 = 0 `

Vì `-3x^2 + 2x - 2 = -3(x^2 - 2/3x - 2/3) = -3[x^2 - 2*x*1/3 + (1/3)^2]-5/3` $\\$ `= -3(x - 1/3)^2 - 5/3 <= -5/3 <0`

=> PT vô nghiệm

=> `S = emptyset`

Bài 2 :

`10/3 - (3a - 1)/(4a + 12) - (7a + 2)/(6a + 18) = 2(a ne -3)` $\\$ `<=> 10/3 - (3a - 1)/[4(a + 3)] - (7a + 2)/[6(a + 3)] = 2` $\\$ `<=> (10*8(a + 3))/(3*8(a + 3)) - [(3a - 1)*6]/[4*6(a + 3)] - [4(7a + 2)]/[6*4(a + 3)] = [2*24(a + 3)]/[24(a + 3)]` $\\$ `<=> [80(a + 3)]/[24(a + 3)] - [(3a - 1)*6]/[24(a + 3)] - [4(7a + 2)]/[24(a + 3)] = [48(a + 3)]/[24(a + 3)]` $\\$ `=> 80(a + 3) - (3a - 1)*6 - 4(7a + 2) = 48(a + 3)` $\\$ `<=> 80a + 240 - 18a + 6- 28a - 8 = 48a + 144` $\\$ `<=> 80a + 240 - 18a + 6 - 28a - 8 - 48a - 144 = 0` $\\$ `<=> -14a + 94 = 0 <=> -7a + 47 = 0` $\\$ `<=>a = 47/7(tmđk)`

Vậy a = `47/7`

Bài 3 :

`((x + 3)/(x- 2))^2 + 6((x - 3)/(x + 2))^2 - 7((x^2 - 9)/(x^2 - 4)) = 0(x ne pm2)` $\\$ `<=> (x+3)^2/(x - 2)^2 + [6(x - 3)^2]/[(x + 2)^2] - [7(x^2 - 9)]/(x^2 - 4) = 0` $\\$ `<=> (x + 3)^2/(x - 2)^2 + [6(x - 3)^2]/[(x + 2)^2] - [7(x + 3)(x - 3)]/[(x - 2)(x + 2)] = 0` $\\$ `<=> [(x + 3)^2(x+2)^2]/[(x - 2)^2(x + 2)^2] + [6(x - 3)^2(x - 2)^2]/[(x - 2)^2(x + 2)^2] - [7(x + 3)(x - 2)(x + 2)(x - 3)]/[(x - 2)^2(x + 2)^2] = 0` $\\$ `<=> (x + 3)^2(x + 2)^2 + 6(x - 3)^2(x - 2)^2 - 7(x^2 - 9)(x^2 - 4) = 0` $\\$ `<=> x^4 + 10x^3 + 37x^2 + 60x + 36 + 6x^4-60x^3+222x^2-360x+216 - 7(x^4 - 4x^2 - 9x^2 + 36) = 0` $\\$ `<=> x^4 + 10x^3 + 37x^2 + 60x + 36 + 6x^4 - 60x^3 + 222x^2 - 360x + 216 - 7x^4 + 28x^2 + 63x^2 - 252 = 0` $\\$ `<=>-50x^3 + 350x^2 - 300x = 0` $\\$ `<=> -50x(x^2 - 7x + 6) = 0` $\\$ `<=> -50x[x^2 - x - 6x + 6] = 0` $\\$ `<=> -50x[x(x - 1) - 6(x - 1)] = 0` $\\$ `<=> -50x(x - 1)(x - 6) = 0 <=> x(x - 1)(x - 6) = 0`

Đến đây tìm nghiệm được rồi

Bài 3 hơi dài bạn thông cảm tí :<