If the absolute value of 3x minus one equals the root six minus two, then what is x equal to? It is better to have a process

If the absolute value of 3x minus one equals the root six minus two, then what is x equal to? It is better to have a process

|3x-1 | = radical 6-2
|3x-1|=2.449-2
|3x-1|=0.449
3x-1=0.449 OR 3x-1=-0.449
3x=1.449 OR 3x=0.551
x=0.483 OR x=0.184

What is the absolute value of the root three minus the absolute value of the root two plus the absolute value of the root three minus two minus the absolute value of the root two minus one?

Original formula = radical 3-radical 2 + 2-radical 2 - (radical 2-1)
=2 + radical 3-2 radical 2-radical 2 + 1
=3 + root sign 3-3 root sign 2

Given that 0 is less than x less than 1, simplify the square of the root sign (x minus 1 / x) minus 4 the square of the root sign (x plus 1 / x) minus 4

The square of the root sign (x minus 1 / x) plus 4 minus the square of the root sign (x plus 1 / x) minus 4
√[(x-1/x)²+4]-√[(x+1/x)²-4]
=√(x²-2+1/x²+4)-√(x²+2+1/x²-4)
=√(x²+2+1/x²)-√(x²-2+1/x²)
=√[(x+1/x)²]-√[(x-1/x)²]
Because the square of the root of 0 (x minus 1 of x) is 4 minus the square of the root (x plus 1 of x) is reduced to 4
The original formula = (x + 1 / x) - (1 / x-x) = x + 1 / X-1 / x + x = 2x
Please take the answer and support me

It is known that X and y are real numbers, and Y ＜ root X-1 + Radix 1-x + 1 / 2, simplification: Y-1 / 1 * Radix (1-y) square

If both √ (x-1) and √ (1-x) are to be meaningful, then x = 1
So y < 1 / 2
Therefore, 1-y > 0
So 1 / 1 of Y-1 * root (1-y) square = - 1

Given the root sign x = (root a) minus 1 / 2 of (root a), simplify the root sign (the square of x plus 4x)

Square on both sides
x=a-2+1/a
x+2=a+1/a
Square on both sides
x²+4x+4=a²+2+1/a²
x²+4x=a²-2+1/a²=(a-1/a)²
So the original formula = √ (A-1 / a) 2
=|a-1/a|

Given a ≠ 0, simplify the square of root 4 + (1 / 1 of A-A)=

√（4+（a-1/a)²）=√（4+（a²-2+1/a²））=√（a²+2+1/a²）=a+1/a

If x satisfies 0.5 less than or equal to x less than or equal to 2, what is the result of simplifying the absolute value (X-2) + the square - X of root x + 1 / 4?

Absolute value (X-2) + the square of root X - x + 1 / 4, should there be a bracket after the root?
=|x-2|+√(x²-x+1/4)
=|x-2|+√(x-1/2)²
=|x-2|+|x-1/2|
=2-x+x-1/2
=3/2

If the square root of (2x-1) of (2x-1) is reduced

1-2X≥0,x≤1/2
x-1≤-1/2
2x-1≤0
The absolute value of the square of root (x-1) plus 2x-1
=1-x+1-2x
=2-3x

If the result of simplifying the absolute value root sign of 1-x (the square of x-8x + 16) is 2x-5, then the value range of X is ditto

2x-5 = (x-1) - (4-x), so | 1-x | - radical (x-8x + 60) = | 1-x) - radical (x-4) = (x-1) - (4-x), we can get X-1 ≥ 0 and x-4 ≤ 0, so x ≥ 1 and X ≤ 4, we get 1 ≤ x ≤ 4, so the value range of X is [1,4]

If the result of reducing the absolute value of 1 minus the square of the root sign (x minus 4) is 2x minus 5, try to judge the value range of X In a hurry

According to the meaning of the question, we can get the following two formulas
1-x-x-4=2x-5
x-1-x-4=2x-5
The solution is x = 1 / 2, or x = 0