If the square of a plus the square of B minus 4A plus 6B plus 13 equals 0, find the square of a plus the square of B

If the square of a plus the square of B minus 4A plus 6B plus 13 equals 0, find the square of a plus the square of B

a²+b²-4a+6b+13=0
a²-4a+b²+6b+4+9=0
a²-4a+4+b²+6b+9=0
(a-2)²+(b+3)²=0,
(a-2)²>=0,(b+3)²>=0,
a-2=0,a=2,
b+3=0,b=-3,
a²+b²=2²+(-3)²=4+9=13.

If a and B are rational numbers and 2a2-2ab + B2 + 4A + 4 = 0, then A2B + AB2 =?
Except for the former 2 of 2A2, 2 of 2Ab and 4 of 4A, all other 2 are square

The original equation is transformed into:
(a^2-2ab+b^2)+(a^2+4a+4)=0
(a－b)^2+(a+2)^2=0
So A-B = 0, a + 2 = 0
That is, a = - 2, B = a = - 2
So a ^ 2B + AB ^ 2 = AB (a + b) = - 2 × (- 2) (- 2-2) = 16

If a ^ 2 + B ^ 2 + 4a-6b = 0, try to find the value of a ^ B
A ^ 2 is the second power of a, B ^ 2 is the second power of B, and a ^ B is the second power of A,
a^2+b^2+4a-6b+13=0

The three sides of a triangle are 4A + 62a-221-a (where a is an integer)

4a+6 + 2a-2 >21-a
21-a + 2a-2 >4a+6
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