If real numbers x and y satisfy X & # 178; = 4-2x, Y & # 178; = 4-2y, find the value of X & # 178; + Y & # 178 Xsquare + ysquare = 4-2x + 4-2y = 8-2 (x + y) Xsquare = 4-2x, ysquare = 4-2y Then x and y are two X + y = - 2 of the equation M & # 178; + 2m-4 = 0, So x square + y square = 8-2 (x + y) = 8-2 * (- 2) = 12 I don't understand how M & # 178; + 2m-4 = 0 comes from

If real numbers x and y satisfy X & # 178; = 4-2x, Y & # 178; = 4-2y, find the value of X & # 178; + Y & # 178 Xsquare + ysquare = 4-2x + 4-2y = 8-2 (x + y) Xsquare = 4-2x, ysquare = 4-2y Then x and y are two X + y = - 2 of the equation M & # 178; + 2m-4 = 0, So x square + y square = 8-2 (x + y) = 8-2 * (- 2) = 12 I don't understand how M & # 178; + 2m-4 = 0 comes from

Answer: x ^ 2 = 4-2x, y ^ 2 = 4-2y obviously, X and y are the two solutions of the equation m ^ 2 = 4-2m, that is, the two solutions of the equation m ^ 2 + 2m-4 = 0. According to Weida's theorem, there are: x + y = - 2XY = - 4, so: x ^ 2 + y ^ 2 = (x + y) ^ 2-2xy = (- 2) ^ 2-2 * (- 4) = 4 + 8 = 12x ^ 2 = 4-2x, y ^ 2 = 4-2y. from these two equations, we can see that X and y can be exchanged, it
It is known that a = 2x and B is a polynomial. When calculating B + A, Xiao Mahu regarded B + A as B divided by a, and the result is the square of X + 1 / 2 x, then B + a=
Wrong number. It is known that a = 2x and B is a polynomial. When calculating B + A, little careless took B + A as B divided by a, and the result is the square of X + 1 / 2 x - 5, then B + a=
A=2x
B÷A=X²+1/2X-5
B=A(X²+1/2X-5)
=2X(X²+1/2X-5)
=2x³+x²-10x
B+A=2x³+x²-10x+2x=2x³+3x²-10x
Merge similar items 3x ^ 2-4 + 2x ^ 2 + 5x-6 + x ^ 2-5x
Original formula = (3 + 2 + 1) xn + (5-5) x + (- 4-6)
=6x²-10
In the equation 4x, where the square is - 1 - x = 0, the unknown is, the term containing the unknown is, and the constant is
Its unknowns are x, the terms with unknowns are 4x & sup2; and - x, and the constant is - 1
Is it 4x & sup2; - 1-x = 0?
The unknown is X
The term with unknowns is quadratic
The constant is - 1
What is unknown is that X has an unknown term 4x - x, and the constant term is - 1
If x 2 + y = 2x-5 and x 2 + y = 2x-5, then x 2 + y is the maximum______ .
From x2-2x + y = 5, we can get: y = 5-x2 + 2x, substituting x + 2Y to get - 2x2 + 5x + 10, let z = - 2x2 + 5x + 10, ∵ in the quadratic function z = - 2x2 + 5x + 10, a = - 2 < 0, ∵ function has the maximum value, that is, Zmax = 4ac − b24a = 4 × (− 2) × 10 − 524 × (− 2) = 1058
If a polynomial is divided by 2x2-1, the quotient is x, and the remainder is X-1, then the polynomial is______ .
According to the meaning of the question: X (2x2-1) + (x-1), = 2x3-x + X-1, = 2x3-1
Remove the brackets and merge the similar items (3x-1) + (2-5x)=
（3x-1）+(2-5x)=3x-1+2-5x=1-2x
（3x-1）+(2-5x)=3X-1+2-5X=(3X-5X)+(2-1)
（3x-1）+(2-5x)
Remove brackets
= 3x - 1 + 2 - 5x
Merge congeners
= 3x - 5x - 1 + 2
[i.e. = (3x - 5x) - (1 - 2)]
= -2x + 1
= 1 - 2x
=3x-1+2-5x=-2x+1
It is known that there is no X term in (x + 2) (x + 3b), then the value of constant B is_____ ? (x is unknown)
,（X+2)(x+3b)
=x^2+(3b+2)x+6b
3b+2=0
b=-2/3
The coefficient of X term of (x + 2) (x + 3b) is 3B + 2, then B should be equal to - 2 / 3
Is it that simple
Try to explain: no matter X and y are any real numbers, the value of X × x + y * y-2x + 2Y + 40 is always positive
(2x + 2) > ^ 2x + 2 + y = 2x-1 + 2x-2 + y = 2x-1 + 2x-2 + y = 2x-2 + y = 2x-1 + 2
To (x-1) ^ 2 + (y + 1) ^ 2 + 38 > = 38
x*x+y*y-2x+2y+40
=(x-1)^2+(y+1)^2+38
(x-1) ^ 2 is greater than or equal to 0
(y + 1) ^ 2 is greater than or equal to 0
So the value of the original formula is always greater than 0 and is always positive
Matching method
(x-10)^2+(y+1)^2+38
Constant > 0
How to write 5x & # 178; - 3xy & # 178; - 2 (2x & # 178; y-xy & # 178;)?
5x²-3xy²-2(2x²y-xy²)
=5x²-3xy²-4x²y+2xy²
=5x²-4x²y+(-3xy²+2xy²)
=5x²-4x²y-xy²
5x²-3xy²-2(2x²y-xy²)
=5x²-3xy²-4x²y+2xy²
=5x²-4x²y+(-3xy²+2xy²)
=5x²-4x²y-xy²
Give me a point
5x²-5xy-4x²y