Given y = (root 2x-1) - (radical 1-2x) = 2x, find the square root of 4x + 5y-3

Given y = (root 2x-1) - (radical 1-2x) = 2x, find the square root of 4x + 5y-3

If 2x-1 ≥ 0, X ≥ 1 / 2
1-2x ≥ 0, X ≤ 1 / 2
So, x = 1 / 2
Substitute x = 1 / 2 to get: y = 0-0 + 1 = 1
So 4x + 5y-3 = 2 + 5-3 = 4
The square root is ± 2

If y = radical X-5 + Radix 5-x + 2, find the square root of 3x + 5Y

From X-5 ≥ 0, 5-x ≤ 0
Results: X ≥ 5, X ≤ 5
If x = 5, then y = 2
The square root of 3x + 5Y
=±√(3x+5y)
=±√25
=±5

Known X − 8 + | y − 17 | 0, find the arithmetic square root of X + y

According to the meaning of the title:
x−8＝0
y−17＝0 ，
The solution is as follows:
x＝8
y＝17 ，
Then x + y = 25,
Then the arithmetic square root of X + y is 5

Given the root sign X-8 + | y-17| = 0, find the arithmetic square root of X + y

Radical X-8 + | y-17| = 0
x-8=0 x=8
y-17=0 y=17
x+y=25
The arithmetic square root of X + y = 5

If y = (quadratic radical X-8) + (quadratic radical 8-x) + 17, then the arithmetic square root of X + y is

In order to make (quadratic radical X-8) and (quadratic radical 8-x) meaningful, then X-8 ≥ 0, 8-x ≥ 0,
So x ≥ 8, and X ≤ 8,
That is, x = 8,
So y = 17,
Then the arithmetic square root of X + y is 5

Given that the real number x, y satisfies the root sign (x + 3y-1) + y 2 + 2Y + 1 = 0, find the square root of 2x-y To process

So (x + 3y-1) + (y + 1) 2 = 0
Because (y + 1) 2
So x + 3y-1 = 0, (y + 1) 2 = 0
The solution is y = - 1, x = 4
The square root of 2x-y = plus or minus 3

Given that real numbers x and y satisfy (√ 2x-3y-1) + | x-2y + 2 = 0, find the square root of 2x - (4 / 5Y)

√(2x-3y-1)≥0
|x-2y+2|≥0
And √ (2x-3y-1) + | x-2y + 2 | = 0
Therefore √ (2x-3y-1) = 0 and | x-2y + 2 | = 0
That is 2x-3y-1 = 0 x-2y + 2 = 0
The solution is x = 8, y = 5
The square root of 2x - (4 / 5Y) = 2 * 8 - √ (4 / (5 * 5)) = 16-2 / 5 = 15.6

Given that X and y satisfy the root sign 2x-3y-1 + absolute value x-2y + 2 = 0, find the square root of 2x-3 / 5Y

2x-3y-1=0
x-2y+2=0
To solve this system of equations
X=8
Y=5
The square root of (2x-3 / 5Y)
=√ 13 or - √ 13

Given the root sign x-2y-3 + | 2x-3y-5 = 0, find the square root of x-8y

Root x-2y-3 + | 2x-3y-5 | = 0,
Absolute value, root sign, greater than or equal to 0
If the sum is 0, all items are 0
So x-2y-3 = 0
2x-3y-5=0
Solve this equation
x=1,y=-1
The square root of x-8y
=±√9
=±3

Known real numbers x, y satisfy X − 2Y − 3 + (2x − 3Y − 5) 2 = 0, find the square root of x-8y

According to the meaning of the title,
x−2y−3＝0①
2x−3y−5＝0② ，
The solution
x＝1
y＝−1 ，
Therefore, x-8y = 1-8 × (- 1) = 9,
So, the square root of x-8y is ± 3