If we know that the cube of Y-1 under the radical is opposite to that of 3-2x under the radical, and the square root of X-Y + 4 is itself, then x () y ()

From (Y-1) + (3-2x) = 0,
∴2x－y＝2
X-Y + 4 = 1
∴x＝5,y＝8.

Known X − y + 3 and If x + y − 1 is opposite to each other, find the square root of (X-Y) 2

Solution
X − y + 3 and
X + y − 1 is opposite to each other,
Qi
x−y+3+
x+y−1=0，
 X-Y + 3 = 0 and X + Y-1 = 0,
X = - 1, y = 2,
∴（x-y）2=（-1-2）2=9，
∵（±3）2=9，
The square root of (X-Y) 2 is equal to ± 3

If the third root sign 3x-7 and the third root sign 3Y + 4 are opposite numbers to each other, the root sign m-2 and the root sign 1-m + n are also opposite numbers to each other. Find the square root and cube of X + m + y + n

Because the third root sign 3x-7 and the third root sign 3Y + 4 are opposite numbers to each other, the positive and negative after the third root sign remains unchanged,
So 3x-7 + 3Y + 4 = 0
X+Y=3
Because the root sign m-2 and the root sign 1-m + n are both greater than 0, the root sign m-2 + root sign 1-m + n is equal to 0,
So {m-2 = 0
{1－M+N=0
M=2,N=1,M+N=3
So x + m + y + n = 9
The square root of 9 is the positive and negative root sign 3, and the cube is 279
That is, the square root of X + m + y + n is positive and negative root sign 3, and the cube is 279

If the absolute value of 3x-y-1 and the root sign 2x + y-4 are the square root of the same number, find the square root of X + 4Y

The square roots of a number are opposite to each other,
（1)|3x-y-1|=3x-y-1,
∴3x-y-1=2x+y-4,
x-2y=-5(1)
(2)|3x-y-1|=-(3x-y-1)=-3x+y+1
∴-3x+y+1=2x+y-4,
-5x = - 5, x = 1,
Substitute (1) y = 2
∴√（x+4y）=±3.

Can you help me? 3x + 4Y root sign 2x-y is the arithmetic square root of 2x-y, and the other square root of 2x-y is - 5. Find the arithmetic square root of X + y

The other square root of 2x-y is - 5, so 2x-y = 25, so 3x + 4Y root is 2x-y = 5, so 3x + 4Y * 5 = 5, namely 3x + 20Y = 5
When the two equations are combined, 2x-y = 25 3x + 20Y = 5, x + y = 440 / 43

If x, y, m are suitable for the relational formula root sign (3x + 5y-3-m) + root sign (2x + 3y-m) = root sign (x + y-2009) + root sign (2009-x-y), find the arithmetic square root of M + 1912

∵x+y-2009≥0 ,2009-x-y≥0
∴x+y=2009 ① x+y-2009=2009-x-y=0
∴√(3x+5y-3-m)+√(2x+3y-m)=0
∴3x+5y-3-m=0②
2x+3y-m=0③
② - 3, x + 2Y = 3 4
④ It was found that - y = - 2006
Substituting ①, x = 4015
x. Substituting y into ③, M = 2012
∴m-1912=100
The arithmetic square root of m-1912 is 10
The arithmetic square root of M + 1912 = 2012 + 1912 = 3924 is the root number 3924 = 6 root sign 109

Given the absolute value of 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
∴x=8
y=17
∴x+y=25
The arithmetic square root of X + y = 5

Given that the real number x, y satisfies the square + root sign x-2y + 2 = 0 of (2x-3y-1), try to find the square root of 2x-5y3y

∵(2x-3y-1)²+√(x-2y+2)=0
According to the meaning of square root
∴2x-3y-1=0…… (1)
x-2y+2=0…… II.
① (2) simultaneous solution
x=8 y=5
So 2x-3y / 5 = 16-3 = 13
Therefore, the square root of 2x-3y / 5 is ±√ 13

Given that the real number x, y satisfies the root sign 2x-3y + 1 + | x-2y + 2 | = 0, find the square root of 2x-3 / 5Y

√(2x-3y+1)+|x-2y+2|=0
∵√(2x-3y+1)>=0,|x-2y+2|>=0
√ (2x-3y + 1) = 0 | x-2y + 2 | = 0 (the sum of two non negative numbers is 0, which can only be obtained by 0 + 0)
2x-3y+1=0 x-2y+2=0
The solution is: x = 4, y = 3
ν (2x-3) / 5Y = 1 / 3, its square root = ± (√ 3) / 3
Note: we don't understand whether 5Y is on the numerator or the denominator

Given that the real number x y satisfies the root sign 2x-3y-1 + absolute value x-2y + 2 = 0, find the square root of 3x-8 / 5Y

According to the meaning of the title:
2x-3y-1=0；
x-2y+2=0；
y-5=0；
y=5；
x=8；
So the square root of 3x-8 / 5Y = ± √ (3 × 8-8) = ± 4;
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If we know that the cube of Y-1 under the radical is opposite to that of 3-2x under the radical, and the square root of X-Y + 4 is itself, then x () y ()