What is the arithmetic square root of root 4

What is the arithmetic square root of root 4

Two

It is known that a is an integer (Radix 17-2), and (B-1) is the arithmetic square root of 9, and the absolute value A-B = b-a

∵ A is (Radix 17-2) take the integer, (B-1) is the arithmetic square root of 9
∴a=2 b-1=3 b=4
ν absolute value A-B = B-A

Given that x.y is reciprocal to each other, C.D is opposite to each other, the absolute value of a is 3, and the arithmetic square root of Z is 5. Find the value of (c + D) (C-D) + XY + radical Z / A

Because X and y are reciprocal of each other, xy = 1
C. D is opposite to each other: C + D = 0
If the arithmetic square root of Z is 5, then z = 25
(c+d)(c-d)+xy＋√z/a
＝0＋1＋√25/3
=1+5√3/3

It is known that X and y are reciprocal to each other, C and D are opposite to each other. The absolute value of a is 3 and the arithmetic square root of Z is 5 Find: 4 (c + D) + XY+ Z The value of A

According to the meaning of the title
xy=1，c+d=0，a=±3，
z=5，
When a = 3
Original formula = 4 × 0 + 1 + 5
3=8
3；
When a = - 3
Original formula = 4 × 0 + 1-5
3=-2
3．

It is known that x.y is reciprocal to each other, C.D is opposite to each other, the absolute value of a is 3, and the arithmetic square root of Z is 5. Find the square root of C 2-D 2 + XY + a fraction of Z

It is known that X. y is reciprocal to each other, xy = 1
c. D is opposite to each other, C + D = 0
The absolute value of a is 3, a = ± 3
The arithmetic square root of Z is 5, z = 25
The square root of Z in C 2 - D 2 + XY + A
=The square root of (c + D) (C-D) + XY + a fraction Z
=0+1±5/3
=8 / 3 or - 2 / 3

Given that the real number Mn satisfies n = root 4-m (is the square of M) + the square of root M-4 fractional line m-2 to find the value of Mn

N=(√（4-M²）+√(M²-4))/(M-2)
It is known that 4-m 2 ≥ 0, M 2 - 4 ≥ 0
So m 2 = 4, and because m-2 is the denominator, it cannot be 0,
So m = - 2 and N = 0. Mn = 0

m. If n is a real number, and the square of M + 9 + the square of (2-N) under the root sign = 6m, then the square of the algebraic formula m - Mn + the square of n

m²-6m+9+|2-n|=0
(m-3)²+|2-n|=0
∵(m-3)²≥0,|2-n|≥0
∴(m-3)²=0,|2-n|=0
That is, M = 3, n = 2
Original formula = 3 * 3-3 * 2 + 2 * 2 = 7

|The square of X-5 | + radical y + 6 + (Z + 8) is 0. There is another way to find the value of 3x + Y-Z + 1: we know that Mn is a real number and | m-radical sign 3 | + root sign n-2 = 0, find n in M The second problem is to find the nth power of M

(1) | X-5 | plus √ (y + 6) plus (Z + 8) ^ 2 = 0
Then it must meet the following requirements:
x-5=0
Y + 6 = 0
Z + 8 = 0
X = 5, y = - 6, z = - 8
3x plus Y-Z plus 1 = 15-6 - (- 8) plus 1 = 18
(2) | m - √ 3 | plus √ (n-2) = 0
Satisfy M = √ 3, n = 2
n/m=2/√3=2√3/3
Supplement: m ^ n = (√ 3) ^ 2 = 3

We know that the fractional part of real number 7 + root flower 19 is m, and the decimal part of real number 11 radical 19 is n. find the arithmetic square root of M + n

The decimal part of the real number 7 + √ 19 is m = √ 19-4,
The decimal part of the real number 11 - √ 19 is n = 5 - √ 19,
The arithmetic square root of M + n = 1

Let A-B be the real root of a-b

Because a and B are real numbers, 5-a = B + 4 under a-5-double root sign,
The root must be greater than or equal to 0, so
a-5>=0,5-a>=0
Namely
a=5,
thus
5-a = B + 4 = 0 under a-5-double root sign
b=-4
therefore
a-b=5-(-4)=9
The arithmetic square root of A-B = 3