Given that ABC is a real number, and the absolute value of A-1 + B + 1 + square + 3 = 0, find the root of the equation AX + BX + C = 0

Given that ABC is a real number, and the absolute value of A-1 + B + 1 + square + 3 = 0, find the root of the equation AX + BX + C = 0

∵ the absolute value of A-1 + B + 1 under the root sign + (c + 3) square = 0
/ / A-1 = 0 B + 1 = 0 C + 3 = 0 (if the sum of non negative numbers is 0, then each non negative number is 0)
∴a=1 b=-1 c=-3
The solution is: x = (1 ±√ 13) / 2

If the square of AX + BX + C = 0 and the real number ABC satisfies 4a-2b + C = 0, then the equation has a root

If B is not equal to 0, then x = - 2.2. A is not equal to 0, then B ^ 2-4ac = 0, from C = 2 (b-2a) into B ^ 2-4ac = 0, we get (b-4a) ^ 2 = 0, get b = 4A, that is, there is C = 4A, replace ax ^ 2 + BX + C = 0, get x ^

Given that ABC is a real number, and the root sign "A-2" + the absolute value of "B + 1" + the square of "C + 3" = 0, find the solution of the equation AX square + BX + C = 0 X is the square of a times X

From the root "A-2" + "B + 1" absolute value + "C + 3" square = 0
All of them are nonnegative, so a = 2, B = - 1, C = - 3
So the equation is 2x? X-3 = 0
Multiply by cross
(x + 1) (2x-3) = 0
So X1 = - 1 or x2 = 3 / 2

Let ABC be a real number not less than 3, then the minimum absolute value of (radical A-2) + (radical B + 1) + [1 - (radical C-1)] is______

∵a≥3,b≥3,c≥3
∴a-2≥1,b+1≥4,c-1≥2
√(a-2)+√(b+1)+|1-√(c-1)|
=√(a-2)+√(b+1)+√(c-1)-1
≥1+2+√2-1
=2+√2

Given the position of real numbers a, B, C on the number axis as shown in the figure, and | a | = | B |, simplify | + | a + B − (c−a)2−2 c2．

According to the position of real numbers a, B and C on the number axis, C < a < 0, b > 0,
According to the meaning of the question, a and B are opposite numbers,
The original formula = - A + 0 - (A-C) + 2C = 3c-2a

The number of the roots of a-C on the square of a-C is reduced a-------b-------0-------c

2 | B | + | B + C | - the square of root (A-C) - 3 the square of cubic root a
=2|b|+|b+c|-|a-c|-a
=-2b+0-(c-a)-a
=-2b-c
=-b-(b+c)
=-b
=c

Given the position of real number ABC on the number axis as shown in the figure, simplify | a | - | A-B | + | C-A | + | b-c| ----a-----------b-------0---------c----

From the graph, C > 0 > b > A. then, a

If the real numbers a and B satisfy a + 1 + A + B = 0 under the radical sign, then what is the value of the algebraic formula A to the power of 2009 + the power of 2010 of B?

√（a+1）+√（a+b）=0
Then a + 1 = 0 and a + B = 0, a = - 1, B = 1
So a ^ 2009 + B ^ 2010 = - 1 + 1 = 0

Given that a = 2-radical 13, find the value of the second power-4a-13 of the algebraic expression a

The answer is (- 14)
Just take a into the equation

It is known that the real number XY satisfies the square of X + 6x + (X-Y + 1) + 9 = 0 Can you work out the values of X and y/

Yes. Because the number under the square and root sign is not less than zero, then x + 3 = 0, X-Y + 1 = 0
X＝-3 Y＝-2