What will be the answer of (x+1) /2 + 2/(x+1) = (x+1) /3 + 3/(x+1) - 5/6?

The answer is x=[0,-7].

Let [math]y=(x+1)[/math]

[math]\dfrac{y}{2}+\dfrac{2}{y}=\dfrac{y}{3}+\dfrac{3}{y}-\dfrac{5}{6}[/math]

[math]\implies 6y\Big(\dfrac{y}{2}+\dfrac{2}{y}\Big)=6y\Big(\dfrac{y}{3}+\dfrac{3}{y}-\dfrac{5}{6}\Big)[/math]

[math]\implies 3y^2+12=2y^2+18-5y[/math]

[math]\implies y^2+5y-6=0[/math]

[math]\implies (y-1)(y+6)=0[/math]

[math]\boxed{y=1}\text{ or }\boxed{y=-6}[/math]

[math]\implies \boxed{\boxed{x+1=1}}\text{ or }\boxed{\boxed{x+1=-6}}[/math]

[math]\boxed{\boxed{\boxed{x=0,\text{ or }-7}}}[/math]

Verify:

If x=0

[math]\dfrac{0+1}{2}+\dfrac{2}{0+1}=\dfrac{0+1}{3}+\dfrac{3}{0+1}-\dfrac{5}{6}[/math]

[math]\implies \dfrac{1}{2}+2=\dfrac{1}{3}+3-\dfrac{5}{6}[/math]

[math]\implies 2+\dfrac{1}{2}=3+\dfrac{2-5}{6}[/math]

[math]\implies 2+\dfrac{1}{2}=3-\dfrac{1}{2}[/math]

[math]2\tfrac{1}{2}=2\tfrac{1}{2}\checkmark[/math]

If x=-7

[math]\dfrac{-7+1}{2}+\dfrac{2}{-7+1}=\dfrac{-7+1}{3}+\dfrac{3}{-7+1}-\dfrac{5}{6}[/math]

[math]\implies -3-\dfrac{1}{3}=-2-\dfrac{1}{2}-\dfrac{5}{6}[/math]

[math]\implies -3-\dfrac{1}{3}-2+\dfrac{-3-5}{6}[/math]

[math]\implies -3-\dfrac{1}{3}-2-\dfrac{6}{6}-\dfrac{2}{6}[/math]

[math]\implies -3-\dfrac{1}{3}=-2-1-\dfrac{1}{3}[/math]

[math]-3\tfrac{1}{3}=-3\tfrac{1}{3}\checkmark[/math]

Thus verified, x=0, or -7.

Q.E.D.