Determine all real values of x in the equation (x-1)^(1/2)+(2-x)^(1/2) =1 ?

We'll impose the constraints of existence of square
roots:

x - 1>=0

x
>= 1

2 -x >=0

x
=<2

The interval of admissible values for x is [1 ;
2].

Now, we'll solve the equation. We'll subtract sqrt(2-x)
both sides:

sqrt(x-1) = 1 -
sqrt(2-x)

We'll raise to square both
sides:

x - 1 = 1 - 2sqrt(2-x)+ 2 -
x

We'll combine like terms:

x
- 1 = 3  -x - 2sqrt(2-x)

We'll subtract 3 - x both
sides:

2x - 4 =
-2sqrt(2-x)

We'll divide by
-2:

2 - x = sqrt(2-x)

We'll
raise to square both sides:

4 - 4x + x^2 = 2 -
x

We'll subtract 2 - x:

x^2 -
3x + 2 = 0

(x-1)(x-2) =
0

x-1=0

x =
1

x-2=0

x
= 2

Since both values belong to the interval
of admissible values, we'll accept them as solutions of the
equation.