Đáp án:
Giải thích các bước giải:
`1/((x+1)(x+2)) + 1/((x+2)(x+3)) + 1/((x+3)(x+4)) = x/((x+1)(x+4))`
ĐKXĐ : \(\left\{ \begin{array}{l}x\ne-1\\x\ne-2\\x\ne-3\\x\ne-4\end{array} \right.\)
`⇔ (1(x+3)(x+4))/((x+1)(x+2)(x+3)(x+4)) + (1(x+1)(x+4))/((x+1)(x+2)(x+3)(x+4)) + (1(x+1)(x+2))/((x+1)(x+2)(x+3)(x+4)) = (x(x+2)(x+3))/((x+1)(x+2)(x+3)(x+4))`
`⇒ (x+3)(x+4) + (x+1)(x+4) + (x+1)(x+2) = x(x+2)(x+3)`
`⇔ x^3 + 5x^2 + 6x = 3x^2 + 15x + 18`
`⇔ x^3+5x^2+6x-18=3x^2+15x`
`⇔ x^3 + 5x^2 - 9x - 18 = 3x^2`
`⇔ x^3 + 2x^2 - 9x - 18 = 0`
`⇔ (x^3+2x^2) + (-9x-18) = 0`
`⇔ -9(x+2) + x^2(x+2) = 0`
`⇔ (x+2)(x^2-9) = 0`
`⇔ (x+2)(x+3)(x-3) = 0`
`⇔`\(\left[ \begin{array}{l}x+2=0\\x+3=0\\x-3=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-2(KTM)\\x=-3(KTM)\\x=3(TM)\end{array} \right.\)
Vậy `x \in {3}`
