It is known that a, B, C are the three sides of the triangle ABC and satisfy the root sign of the third power It is known that a, B, C are the three sides of the triangle ABC and satisfy the cubic root sign a = 2 and (b-2c + k) ^ 2 + root a-b-2 = 0. If the triangle ABC is an isosceles triangle, find the value of K and the circumference of triangle ABC Please! Boss, can do all write, I am a grade one little bookboy, early used up

If the root of the cubic is a = 2, we can get a = 8
(b-2c + k) ^ 2 + a-b-2 = 0 is the sum of two nonnegative numbers, so each one is equal to o
B-2c + k = 0, a-b-2 = 0; b = 6
If the triangle ABC is an isosceles triangle, C = 8 or C = 6, k = 10 or K = 6, and the circumference C = a + B + C = 22 or 20

A = (root 3-2) to the power of 2011, B = (root 3 + 2) to the power of 2012, and calculate the value of a × B

Because a = (√ 3-2) ^ 2011, B = (√ 3 + 2) ^ 2012 ^ is the power
therefore
A×B
=[(√3-2)^2011]×[(√3+2)^2012]
=[(√3-2)^2011]×[(√3+2)^2011]×(√3+2)
={[(√3-2)(√3+2)]^2011}×(√3+2)
=[(3-4)^2011]×(√3+2)
=[(-1)^2011]×(√3+2)
=-(√3+2)
= -√3-2

Calculation: the root 18 minus (3 minus) the 0 power minus the negative 1 power of (3 points 1)

3 root numbers 2-4

Radical 18-9 / 2-3 + 6 + 0-2 + 2-2

When you type in the future, you'd better use parentheses more often

Given the radical y-2x + [(the absolute value of the square of X + 25)] and the root (5-x) = 0, find the cube root of 7 (x + y) - 20 However, when x = 5, the denominator is zero

If two nonnegative monomials are added together and equal to 0, then there is only one case, that is, both sets of monomials are 0, so y = 2x 5-x = 0
So x = 5, y = 10

Let a = radical B-4 + Radix 4-b divided by the square of B-2 + Half B, and the absolute value of X + y-6 = - radical X-Y + 2, find the value of cube root of ABXY

A = root B-4 + Radix 4-b divided by B-2 + the square of half B
Therefore, B-4 ≥ 0, 4-b ≥ 0
So, B = 4
a=0+0+(4/2)^2=4
The absolute value of X + y-6 = - radical X-Y + 2
x+y-6=0
x-y+2=0
By solving the equations, the following results are obtained
x=2,y=4
Therefore, cube root of ABXY = (4 * 4 * 2 * 4) cube root = 4 * (cube root of 2)

Let a, B satisfy B = ((radical A-4) + (radical 4-A 2) + 4) / (A-2). Find the value of radical (a-2b) + ab

From the meaning of the title, we can get the following: a 2 - 4 = 4 - a 2 = 0 and a ≠ 2
Then the solution is: a = - 2
Then: B = 4 / (- 2-2) = - 1
So:
Radical (a-2b) + radical (AB)
=Radical (- 2 + 2) + radical [- 2 * (- 1)]
=Radical 2

Given that the absolute value of a + b square + (root sign term C-1) - 1 is equal to 4 times of the root term A-2 + 2b-3, find a + 2b-1 / 2C emergency

The original formula can be changed into [(A-2) - 4 √ (A-2) + 4] + (b ^ 2-2b + 1) + | (C-1) - 1-1 | = 0 [(A-2) - 2] ^ 2 + (B-1) ^ 2 + (2 + (A-2) - 2] ^ 2 + (B-1) ^ 2 + 2 + (C-1) - 1-1-1 = 0 because every item is not negative, because every item is not negative, then √ (A-2) - 2 = 0, B-1 = 0, C-1) - 1 = 0, so: a = 6, B = 1, C = 1, C = 2A + 2b-1 / 2C = 6 + 2 + 2 + 1 / (2 * 2 * 2 * 2 * 2 * 2 * 2 * 1 / 2C = 6 + 2-1 2) = 31 / 4

2 times the absolute value of A-1 + 2B + C under the root sign plus the square - C + 1 / 4 of C equals to a + B + C of 0=

2|a-1|+√(2b+c)+(c-1/2)²=0
Absolute value, arithmetic square root, square are greater than or equal to 0
If one is greater than 0, then at least one is less than 0
So all three equations are equal to zero
So A-1 = 0,2b + C = 0, C-1 / 2 = 0
a=1,c=1/2,b=-c/2=-1/4
So a + B + C = 1-1 / 4 + 1 / 2 = 5 / 4

Given that the real numbers a, B and C satisfy the absolute value of A-B + the square of 2B + C + C - 1 / 4 = 0 under the root sign of A-B, find the value of a + B + C

It is known that a, B, C satisfies | A-B | + + (2B + C) + C ^ 2-C + 1 / 4 = 0, that is, A-B | + (2B + C) + (C-1 / 2) ^ 2 = 0, because the three are non negative, add up to be equal to 0, then we can only take 0, so we can all take 0, so | A-B = 0, 2b + C = 0, C-1 / 2 = 0, so a = - 1 / 4, B = - 1 / 4, C = 1 / 2, so a + B + C = (- 1 / 4) + (- 1 / 4), C = 1 / 2, so a + B + C = C = (- 1 / 4) + (- 1 / 4) + (- 1 / 4) + (- 4) + (1 / 2) = 0