If the squares of | B-1 | and (a + 3) are opposite numbers to each other and x = - 1 is the solution of the equation (a + x) / 4-3y = half x + B about X, find the value of square-3 of 2Y

If the squares of | B-1 | and (a + 3) are opposite numbers to each other and x = - 1 is the solution of the equation (a + x) / 4-3y = half x + B about X, find the value of square-3 of 2Y

|The square of b-1|and (a + 3) is opposite to each other
|The square of B-1 | + (a + 3) is 0
|B-1 | = 0, the square of (a + 3) = 0
a=-3
B=1
A = - 3, B = 1, x = - 1 are substituted into the equation
[-3+(-1)]/4-3y=(-1)/2+1
-1-3y=1/2
3y=-3/2
y=-1/2
The square of 2y-3
=The square of 2 * (- 1 / 2) - 3
=2*1/4-3
=1/2-3
=-5/2

Simplification evaluation: 1 / 7 (- 2x + 3Y) (- 2x-3y) - 10 times the square of (x + y), where x = 1.25, y = -0.25 We should use the method of simplification, which should be simple and convenient,

1 / 7 (4x square - 9y Square) - 10 times (x + y)
X + y = 1, so if you remove the back and multiply by (x + y), you get 1 / 7 (4x square - 9y Square) - 10
Then the values of 13 / 16-10 = - 147 / 16 are obtained

Simplification: (2x + 3Y) 2 - (2x + y) (2x-y)

Original formula = 4x2 + 12xy + 9y2 - (4x2-y2)
=4x2+12xy+9y2-4x2+y2
=12xy+10y2．

The square of (x + 3Y) + (2x + y) (X-Y) simplification

(x+3y)²+(2x+y)(x-y)
　　=x²+6xy+9y²+2x²-2xy+xy-y²
　　=3x²+5xy+8y²


Simplify evaluation (2x + 3Y) squared - (2x + y) (2x-y)

（2X+3Y）²-（2X+Y）（2X-Y）
=4X²+12XY+9Y²-（4X²-Y²）
=4X²+12XY+9Y²-4X²+Y²
=12XY+10Y²

Simplification evaluation: 3x's Square y-3y's Square x + 2XY's Square - 2x's Square y, where x = 1 / 2, y = negative 1

3x²y-3y²x+2xy²-2x²y
=3x²y-2x²y-3xy²+2xy²
=x²y-xy²
=xy(x-y)
=1/2*(-1)*[1/2-(-1)]
=-1/2*3/2
=-3/4

If the square of (2x + 3y-6) is opposite to the absolute value of (3x-2y + 17), then x =? Y =? Is satisfied? I only know the answer, but I don't know how to calculate it,

If they are opposite to each other, the sum is equal to 0
So (2x + 3y-6) 2 + | 3x-2y + 17| = 0
If one is greater than 0, the other is less than 0
So both equations are equal to zero
So 2x + 3y-6 = 0
3x-2y+17=0
X = - 3, y = 4

The square of rock | X - 2Y + 3 | and (2x + 3y-10) are opposite to each other. Try to find the value of X + y

If the squares of two numbers are opposite to each other, then both numbers are 0
| X -2Y+3 |=0
（2X+3Y-10）=0
Calculate y = 16, x = 29
X+Y=45

If the square of (2x-3y-6) and the absolute value of 3x-2y + 17 are opposite to each other, X_ Y_

Because the square of (2x-3y-6) is greater than or equal to 0, the absolute value of 3x-2y + 17 is also greater than or equal to 0
Because the values of the two formulas are opposite, they can only be equal and both are 0
The results show that x = - 63 / 5
y=-52/5

Given: the absolute value of x-3 / 2 + (y + 2) square = 0, find the value of X-2 (1 / 4x-1 / 3Y's Square) + - 3 / 2x + 1 / 3Y's Square)

Absolute value of x-3 / 2 + (y + 2) square = 0
x-3/2=0 y+2=0
x=3/2 y=-2
X-2 (1 / 4x-1 / 3Y squared) + - 3 / 2x + 1 / 3Y squared)
=x-1/2x+2/3y^2-3/2x+1/3y^2
=-x+y^2
=-3/2+(-2)^2
=-3/2+4
=5/2