Môn Toán Lớp 8: `X+2/x+1+3/x-2=3/x^2-x-2+1` tìm `x`.

Giải đáp:

 \frac{x+2}{x+1}+\frac{3}{x-2}=\frac{3}{x^{2}-x-2}+1 (đk: x\ne-1;x\ne2)

⇔ \frac{x+2}{x+1}+\frac{3}{x-2}=\frac{3}{x^{2}-2x+x-2}+1

⇔ \frac{x+2}{x+1}+\frac{3}{x-2}=\frac{3}{x.(x-2)+1.(x-2)}+1

⇔ \frac{x+2}{x+1}+\frac{3}{x-2}=\frac{3}{(x+1).(x-2)}+1

⇔ \frac{(x+2).(x-2)}{(x+1).(x-2)}+\frac{3.(x+1)}{(x+1).(x-2)}=\frac{3}{(x+1).(x-2)}+\frac{(x+1).(x-2)}{(x+1).(x-2)}

⇔\frac{x^{2}-4}{(x+1).(x-2)}+\frac{3x+3}{(x+1).(x-2)}=\frac{3}{(x+1).(x-2)}+\frac{x^{2}-x-2}{(x+1).(x-2)}

⇒ x^{2}-4+3x+3=3+x^{2}-x-2

⇔ x^{2}+3x+(3-4)=x^{2}-x+(3-2)

⇔ x^{2}+3x-1=x^{2}-x+1

⇔ x^{2}+3x-1-x^{2}+x-1=0

⇔(x^{2}-x^{2})+(3x+x)-(1+1)=0

⇔4x-2=0

⇔4x=2

⇔x=\frac{2}{4}

⇔x=\frac{1}{2} (tmđk)

Vạy x=\frac{1}{2}