`5/(x^2) = 1 + 4/((x-2)^2)`
`ĐKXĐ: x ne 0; x ne 2`
`Pt <=> 5/(x^2) - 1 - 4/((x-2)^2) = 0`
`<=> (5(x-2)^2 - x^2 (x-2)^2 - 4x^2)/(x^2 (x-2)^2) = 0`
`<=> 5(x-2)^2 - x^2 (x-2)^2 - 4x^2 = 0`
`<=> 5(x^2 - 4x + 4) - x^2 (x^2 - 4x + 4) - 4x^2 = 0`
`<=> 5x^2 - 20x + 20 - x^4 + 4x^3 - 4x^2 - 4x^2 = 0`
`<=> -3x^2 - 20x + 20 - x^4 + 4x^3 = 0`
`<=> x^4 - 4x^3 + 3x^2 + 20x - 20 = 0`
`<=> (x^4 - x^3) - (3x^3 - 3x^2) + (20x - 20) = 0`
`<=> x^3(x - 1) - 3x^2 (x - 1) + 20 (x - 1) = 0`
`<=> ( x^3 - 3x^2 + 20)(x - 1) = 0`
`<=> (x^3 + 2x^2 - 5x^2 - 10x^2 + 10x^2 + 20)(x - 1) = 0`
`<=> [(x^3 + 2x^2) - (5x^2 + 10x) + (10x + 20)](x - 1) = 0`
`<=> [x^2(x +2) - 5x(x + 2) + 10(x + 2)](x - 1) = 0`
`<=> (x^2- 5x + 10)(x + 2)(x - 1) = 0`
Do `x^2 - 5x + 10 = x^2 - 2x . 5/2 + (5/2)^2 + 15/4 = (x - 5/2)^2 + 15/4 >0`
Nên: `x + 2 = 0` hoặc `x - 1 = 0`
`<=> x = -2` hoặc `x = 1` (Thỏa mãn)
Vậy ....
