Given that a, B, C are the three side lengths of △ ABC, we try to simplify it as follows: root sign (a + B + C) 2 + radical sign (a-b-c) 2 + | b-a-c|

Given that a, B, C are the three side lengths of △ ABC, we try to simplify it as follows: root sign (a + B + C) 2 + radical sign (a-b-c) 2 + | b-a-c|

a. B, C are the three sides of △ ABC
Then: a + B + C > 0
aa-b-c<0
bb-a-c<0
Root sign (a + B + C) 2 + root sign (a-b-c) 2 + | b-a-c|
=a+b+c+b+c-a+a+c-b
=a+b+3c

Let a be the integer part of Radix 17, and B-1 be the arithmetic square root of absolute value of 2. Find the root sign (a + b)

A is the integral part of Radix 17, that is, a = 4
B-1 is the arithmetic square root of the absolute value of 2, that is, B-1 = root 2, then B = 1 + root 2
Root number (a + b) = root number (4 + 1 + root number 2) = root number (5 + root number 2)

It is known that the arithmetic square root of 2a-1 is 3, the arithmetic square root of 3A + B-1 is 4, and C is the integral part of root 13. Find the arithmetic square root of a + 2b-c

2a-1 = 9, the solution is: a = 5
3A + B-1 = 16, B = 2
√ 13 is between 3 and 4, so C = 3
∴a+2b-c=6
The arithmetic square root of a + 2b-c is √ 6

(1) A-b-1 is known|+ 2A + B − 5 = 0, what is the arithmetic square root of AB? (2) If y= 1−x+ X − 1 + 2, then what is the arithmetic square root of 2x + y?

（1）∵|a-b-1|+
2a+b-5=0，
Qi
a-b-1=0
2a+b-5=0 ，
The solution is as follows:
A=2
b=1 ，
∴ab=21=2，
The arithmetic square root of AB is
2；
（2）∵y=
1-x+
x-1+2，
Qi
1-x≥0
x-1≥0 ，
The solution is: x = 1,
∴y=2，
∴2x+y=4，
The arithmetic square root of 2x + y is: 2

Given that the absolute values of the root sign x-3 and y + 1 are opposite numbers to each other, find the arithmetic square root of the square root of the square of the square root of the square of the square root of the square of the square of the square root of the square of the square root of the square of the square root of the square of the square root of the square of the square root of the square

It is known that the absolute values of the radical x-3 and y + 1 are opposite to each other
Radical (x-3) + |y + 1| = 0
x-3=0,y+1=0
By solving the equation, x = 3, y = - 1
The arithmetic square root of the power of 9y to the power of x = root [9 * (- 1) ^ 2010-3 ^ 2] = 0

Given that y = radical X-2 + Radix 2-x + 2, try to find the arithmetic square root of y to the power of X

Because the number under the root sign is greater than or equal to 0
So x - 2 ≥ 0 and 2 - x ≥ 0
So x = 2
So y = 0 + 0 + 2 = 2
So the x power of y = 2? = 4
So the arithmetic square root of y to the power of X is two

Given that A-B is a real number and a-5 minus 3 times under the root sign, 10-2a = B-4, find the arithmetic square root of a to the B power

√(a-5)-3√(10-2a)=b-4
Make the radical meaningful
Then a-5 ≥ 0
10-2a≥0
The solution is 5 ≤ a ≤ 5, that is, a = 5
∴a-5=10-2a=0
b-4=0+0=0,b=4
∴√(a^b)=√(5^4)=5²=25

a. B is a real number, (Radix a-5) - 2, Radix 10-2a = B-4, find the arithmetic square root of a-b emergency

Make the range of the root of a meaningful
a-5≥0
10-2a≥0
A=5
Bring in B = 4
A-B = 1, the arithmetic root is 1

Given that real numbers a and B satisfy | A-2 | + root sign B-8 = 0, find the arithmetic square root of a * B

a-2=0
b-8=0
∴a=2
B=8
∴√(ab)=√ 16=4

What is the arithmetic square root of root 2A + 3? Why?

Under the root sign, the root sign 2A + 3 is the square root of this number. Take the positive root. It should be meaningful a > = - 3 / 2