Given the absolute value of a + 1 / 3 + the square of 2B + 1 + (C-2) under the root sign = 0, find the BC power of A It's 2B + 1 under the radical Given that a, B are rational numbers, and satisfy that under 5a-2b times the root sign, six is equal to the square of (root 3 + root 2), and find the 2014 power of (AB)

If the sum of three nonnegative numbers is zero, then they are zero respectively
A + 1 / 3 = 0, a = - 1 / 3
2B + 1 = 0, B = - 1 / 2
c-2=0,C=2
therefore
A ^ BC = (- 1 / 3) ^ - 1
=-3

Given the set a = {- A, A2, AB + 1} and B = {- 3} A3 ，a a. If the elements in 2B} are the same, find the values of real numbers a and B

It is known that a = {- A, a, AB + 1}, B = {- A, 1, 2b},
∵ the elements a and B are the same,
Qi
a＝1
AB + 1 = 2B or
a＝2b
ab+1＝1 ，
The solution
a＝1
B = 1 or
a＝0
B = 0 is not consistent with the meaning of the question, so it is omitted;
So the answer is: a = 1, B = 1

Given that the absolute value of a = 2, the absolute value of B = root 2, the angle between a and B is 45 degrees, and λ B-A is perpendicular to a, the value of real number λ is calculated

(λb-a)a=cos45× λ｜a｜｜b｜-a^2=√2/2×λ｜a｜｜b｜-｜a｜^2=0
λ=2

Absolute value a-2010 + Radix a-2011 = a, calculate the value of a-2010 square (is the square of 2010)

Because the radical (a-2011) is meaningful,
So a > 2011,
So | a-2010 | = a-2010,
So a-2010 + Radix (a-2011) = a,
Root number (a-2011) = 2010,
Both sides are square at the same time, a-2011 = (2010 ^ 2),
So a - (2010 ^ 2) = 2011

The absolute value of 2009 minus a plus the root sign a minus 2010 is equal to A. find the value of a minus 2009 square

The absolute value of 2009 minus a plus the root a minus 2010 equals a,
a-2010>=0
a>=2010
|2009-a|+√a-2010=a
a-2009+√a-2010=a
√a-2010=2009
a-2010=2009^2
a-2009^2=2010
Find a minus 2009 square

Let a be equal to the square of root TWO-1, B = - 3, and C = - the absolute value of negative root two. Compare the size of ABC

If √ = - 2, C = √ - 2
b〈a〈c

3 √ 2 ＋| 3-3 √ 2 | - √ (- 5) 2. The absolute value of 3 times root 2 plus 3 minus 3 times root 2 minus the square of root sign (minus 5) is equal to (=)?

solution
simple form
=3√2+(3√2-3)-5
=6√2-8

It is known that the three sides of the triangle ABC are ABC respectively, and ABC satisfies the absolute value of the root sign A-3 + B-4 + the square of c-6c + 9 = 0. Try to judge the shape of ABC and explain the reason, and find the circumference and area of ABC

The absolute value of the root (A-3) + (B-4) + the square of (C-3) = 0, all of which are greater than or equal to 0
So all three of them are equal to zero
A=3
B=4
C=3
right triangle
Perimeter 3 + 4 + 5 = 16
Area 3 * 4 / 2 = 6

Given that the triangle a ` B ` C of the triangle ABC satisfies the absolute value of a ^ 2 + B + [(radical C-1) - 2] = 6A + 2 * (radical B-3) - 7), try to judge the shape of triangle ABC

a^2+b+|√(c-1)-2|=6a+2*√(b-3)-7
(a^2-6a+9)+[(b-3)-2√(b-3)+1]+|√(c-1)-2|=0
(a-3)^2+[√(b-3)-1]^2+|√(c-1)-2|=0
The absolute value of sum of squares is greater than or equal to 0, and the sum is equal to 0. If one of them is greater than 0, at least one of them is less than 0, which is not true
So all three are equal to zero
So (0, - 3) - 1
A=3
√(b-3)=1
√(c-1)=2
b=4,c=5
a^2+b^2=c^2
So it's a right triangle

Given that the rational number a satisfies │ 2007-a │ + Radix [A-2008] = a, then what is the square value of A-2007? 11

│ 2007-a │ + radical [A-2008] = a
We can know that A-2008 > = 0, so a > = 2008
So the original equation becomes
a-2007+√(a-2008)=a
√(a-2008)=2007
a=2007^2+2008
a-2007^2
=2007^2+2008-2007^2
=2008