Given x = 2-radical 3, y = 2 + radical 3, find the value of the algebraic formula x square + y square

Given x = 2-radical 3, y = 2 + radical 3, find the value of the algebraic formula x square + y square

x²+y²
=x²+2xy+y²-2xy
=(x+y)²-2xy
=(2-√3+2+√3)²-2(2-√3)(2+√3)
=4²-2×(4-3)
=16-2
=14

Given that the square of the sum of 2A + B-4 plus 4a-b-2 under the root sign is equal to 0, find the value of one-third of the algebraic expression multiplied by the square of - 2Ab square

√(2a+b-4)+√(4a-b-2)=0
√(2a+b-4)=0,√(4a-b-2)=0
2a+b-4=0
4a-b-2=0
The solution
A=1
B=2
1/3*(-2ab²)²
=1/3*(-2*1*2²)²
=1/3*(-8)²
=64/3

Given that a, B, C are unequal positive real numbers, and cut ABC = 1, it is proved that radical a + radical B + radical C

1 / A + 1 / b > = 2 times root number (1 / AB) root number C = root number (1 / AB)
So 1 / A + 1 / b > = 2 times the root number C 1 / B + 1 / C > = 2 times the root number a 1 / C + 1 / a > = 2 times the root number B
1 / A + 1 / B + 1 / C > = root a + root B + root c
So the condition of equal sign is a = b = C
And a, B, C are positive numbers that are not equal to each other
So: (1 / A + 1 / B + 1 / C) > root a + root B + root C

Given the real number a.b.c. and a + B + C = 2 (radical a + radical (B-1) + radical (C-2)), calculate the ABC value

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Given that a and B are real numbers, and b square + root A-4 + 9 = 6B, if a and B are two sides of △ ABC, we can find the value range of the third side C We know that a and B are real numbers, and satisfy b square + root A-4 (+ 9 outside the root) + 9 = 6B. If a and B are both sides of △ ABC, we can find the value range of the third side C

∵b²+√（a-4)+9=6b
∴b²-6b+9+√(a-4)=0
That is (B-3) 2 + √ (A-4) = 0
∴b=3,a=4
The value range of the third side C is 1

a. B, C are real numbers, and (2-A) 2 + radical a 2 + B + C + C + 8 = 0, ax 2 + BX + C = 0, find the algebraic formula 3x 2 + 6x + 1

a=2,c=-8,4+b-8=0,b=4
2x^2+4x-8=0,x^2+2x-4=0
3x^2+6x+1=3(x^2+2x)+1=3*4+1=13

Let a, B and C be real numbers, and satisfy (2-A) 2 + (a 2 + B + C) + | C + 8 | = 0, ax? 2 + BX + C = 0, Find the arithmetical square root of the formula x 2 + 2x

The sum of the three nonnegative numbers (2-A) 2, (a 2 + B + C) and | C + 8 | is 0, and all of them must be 0
2-A = 0, a 2 + B + C = 0, and C + 8 = 0,
The solution is: a = 2, B = 4, C = - 8
Ax 2 + BX + C = 0 becomes 2x 2 + 4x-8 = 0
So 2x 2 + 4x = 8
And the arithmetic square root of 8 is 2 √ 2
So the arithmetic square root of the formula x? 2 + 2x is 2 √ 2
[if you don't know, ask again; if you are satisfied, I wish you good luck

a. B, C satisfy (2-A) 2 + radical sign (a 2 - B + C) + C + 8 = 0, and ax 2 + BX + C = 0, find the value of the algebraic formula x? - 2x + 5

∵（2-a）²+√（a²-b+c）+|c+8|=0
And ∵ (2-A) 2 ≥ 0 a ∵ - B + C ≥ 0, C + 8 ≥ 0
∴2-a=0 a²-b+c=0 c+8=0
A = 2, B = - 4, C = - 8
That is, the original equation is
2x²-4x-8=0
We get x? 2x = 4
If x 2 - 2x + 5 is brought in, the
Original formula = 4 + 5 = 9

Let ABC be real numbers and satisfy (2-A) + (a + B + C) + | C + 8 | = 0 ax + BX + C = 0

According to (2-A) + root sign (a + B + C) + | C + 8 | = 0
2-a=0,a^2+b+c=0,c+8=0
A = 2, C = - 8, B = 4
Substituting ax ^ 2 + BX + C = 0
We get 2x ^ 2 + 4x-8 = 0
x^2+2x-4=0
That is, x ^ 2 + 2x = 4
So √ (x ^ 2 + 2x) ^ 2 = √ 16 = 4

It is known that ABC is a real number and the absolute value of the root sign A-2 + B + 1 + (c + 3) 2 = 0, ax 2 + X + C = 0

Because the absolute values of the roots A-2, B + 1, (c + 3) 2 are nonnegative, and the sum is 0,
So A-2 = 0, B + 1 = 0, C + 3 = 0,
That is, a = 2, B = - 1, C = - 3
So the equation AX? 2 + X + C = 0 becomes: 2x? 2 + x-3 = 0,
The solution is (2x + 3) (x-1) = 0
So X1 = - 3 / 2, X2 = 1