Rationalizing may help

A different solution is as follows:

Let $a^2=x+1$ and $b^2=x-1$ .

We see that $a^2-b^2=2$ . Also, from the equation, $a-b=1$ .

Dividing the second equation from the first we get $a+b=2$ .

Adding this to the second equation and simplifying, we get $a=\dfrac{3}{2}$ so $b=\dfrac{1}{2}$ .

Thus $x-1=\dfrac{1}{4}$ so $x=\dfrac{5}{4}=\boxed{1.25}$ .

Squaring and rearranging gives $2x-1=2 \sqrt{x^2-1}$ . Squaring again, canceling and rearranging gives $4x-5=0$ , that is, $x=5/4=1.25$ . It's easy to check that this value indeed works.

First,we change the equation to $\sqrt{x+1} - \sqrt{x-1} = 1$ Then,we rearrange the terms to get $\sqrt{x+1} - 1 = \sqrt{x-1}$ Squaring both sides,we get $x + 1 - 2\sqrt{x+1} + 1 = x -1$ Subtracting the $x$ from both sides,we get $2 - 2\sqrt{x+1} = -1$ Then,we get $2\sqrt{x+1} = 3$ .Dividing by 2,we get $\sqrt{x+1} = \frac{3}{2}$ .Squaring both sides,we get $(x+1) = \frac{9}{4}$ So, $x = \frac{5}{4} = 1.25$

A common solution is as follows:

{ (x+1) }^{ 1/2 }=1+{ (x-1) }^{ 1/2 }

squaring both sides to the above equation,

(x+1)=1+x-1+2{(x-1)}^{1/2}

1=2{(x-1)}^{1/2}

again squaring the both sides to the above equation,

1=4(x-1)

x=5/4.

Hey yo,

(x+1)^0.5 = 1 + ( x - 1)^0.5

Squaring both sides,

1 = 2(x-1)^(0.5)

(x-1)^(0.5) = 1/2,

Squaring both sides once again,

x = 1/4 + 1 = 1.25.....(try to insert the value to check the answers)

(1.25 + 1)^(0.5) - ( 1.25 -1 )^(0.5) = 1 (yeah x = 1.25)....

thanks.....