Given the position of a, B, C on the number axis as shown in the figure, simplify | a|- (a+c)2+ (c−a)2+ (c−a)2-（ b）2．

Given the position of a, B, C on the number axis as shown in the figure, simplify | a|- (a+c)2+ (c−a)2+ (c−a)2-（ b）2．

According to the meaning of the title: C < 0 < a < B, and | a | C |,
∴a+c＜0，c-a＜0，
Then the original formula = a + A + C + a-c + A-C-B = 4a-c-b

The edges to which the inner angles of △ ABC are opposite are a B C, asinasin B + bcos? A = radical two a, then B / a =?

asinAsinB+bcos²A=√2a
Sine theorem
sin²AsinB+sinBcos²A=√2sinA
sinB(sin²A+cos²A)=√2sinA
sinB=√2sinA
Sine theorem
b=√2a
∴b/a=√2

If the opposite sides of the three inner angles a, B and C of △ ABC are a, B, C, asinasinasinb + bcos? A = radical 2 × a, then B / A is equal to?

√2

In the triangle ABC, the opposite sides of the inner angle A.B.C are a, B, C respectively. We know that B = C, 2b = (radical 3) A. --- find cos (2a + 45) In the triangle ABC, the opposite sides of the inner angle A.B.C are a, B, C respectively. We know that B = C, 2b = (radical 3) a ----Find cos (2a + 45 degrees)

B=C ；2b=√3a；∴ b = √3/2 asin(A/2) = (a/2) / b = √3/3cos(A/2)= √（1-1/3）= √6/3sinA = 2sin(A/2)cos(A/2) = 2√2/3cosA = √（1-8/9）=1/3sin2A = 2sinAcosA = 4√2/9cos2A = -√（1-32/81）= - 7/9cos（...

Given that a is the integral part of Radix 170 and B-3 is the arithmetic square root of 400, find the root a + B

13<√170<14
∴a=13
b-3=√400=20
b=23
a+b=13+23=36

If the absolute value of the square - 9 of (a-4b) + B under the root sign = 0, find the square root of 21-a-b under the root sign

[√﹙a－4b﹚/√﹙7－a﹚]²＋|b²－9|＝0
b²－9＝0
b＝±3
a－4b＝0
a＝4b＝±12
∵ 7－a＞0.
∴ a＜7
∴ a＝﹣12, b＝﹣3
±√﹙21－a－b﹚＝±√﹙21＋12＋3﹚＝±6.

Given that the arithmetic square root of 2A + 8 is 4, the square root of 3A + B-1 is plus or minus 3, and C is the integral part of 13 under the root sign. Find the square root of 3a-2b-c ^ 2

It is known that the arithmetic square root of 2A + 8 is 4
Then 2A + 8 = 4 * 4
A=4
The square root of 3A + B-1 is plus or minus 3,
We can see that 3A + B-1 = 3 ^ 2
12+b-1=9
b=-2
C is the integer part of 13 under the radical, then C is equal to 3
It can be seen that 3a-2b-c ^ 2 = 3 * 4 + 2 * 2-3 * 3
=7
The square root of 3a-2b-c ^ 2 is ±√ 7

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

The arithmetic square root of 2a-1 is 3, so 2a-1 = 9, a = 5
The square root of 3a-b-1 is plus or minus 4, so 3a-b-1 = 16, B = - 2
C is the integer part of root 13, so C = 3
a+2b+c+4=5+2*(-2)+3+4=8
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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 I already have the answer, but the question is, why should X-2 and 2-x be greater than or equal to zero?

If the number under the root sign is greater than or equal to 0, that is, the number under the root sign must be 0 or positive, otherwise it does not hold

If y = root sign (x-3) + root sign (3-x) + 2, then the arithmetic square root of - x power of Y

Y = radical (x-3) + radical (3-x) + 2
Then x-3 ≥ 0, 3-x ≥ 0
Then x ≥ 3, X ≤ 3
So x = 3
Then y = 0 + 0 + 2 = 2
The - x power arithmetic square root of y = the arithmetic square root of - 3 power of 2 = the arithmetic square root of 1 / 8 = 4 √ 2