Given that X and Y belong to R, and xy = 2, then the value range of X + y is given

Given that X and Y belong to R, and xy = 2, then the value range of X + y is given

Let x + y = a
y=a-x
Substituting xy = 2
x²-ax+2=0
If x is a real number, then △≥ 0
a²-8≥0
a²≥8
therefore
x+y≤-2√2,x+y≥2√2

Given that x.y belongs to R, x ^ 2 + y ^ 2-xy is less than or equal to 1, find the value range of X + y
Hope to get a variety of solutions

1)(x^2+y^2-xy)'=(1)'
There are 2x + 2yy '- y-xy' = 0
Let y '= - 1
There are 2x-2y-y + x = 0
There is x = y
x=y=+_1
-2

If xsquare - XY + ysquare = 1, then the value range of xsquare - ysquare

From X & # 178; - XY + Y & # 178; = 1, we get X & # 178; - XY + Y & # 178; - 1 = 0
If x is regarded as an unknown, then the discriminant Y & # 178; - 4 (Y & # 178; - 1) ≥ 0, 0 ≤ Y & # 178; ≤ 4 / 3 is the same as 0 ≤ X & # 178; ≤ 4 / 3
-4/3≤x²-y²≤4/3

If x square plus XY plus y square = 1, find the value range of the square of X - XY + y square?

Because x ^ 2 + XY + y ^ 2 = 1
So: 1 = x ^ 2 + y ^ 2 + XY ≥ 2XY + xy = 3xy
That is: XY ≤ 1 / 3
x^2-xy+y^2=(x^2+y^2)-xy=(1-xy)-xy=1-2xy≥1-2*1/3=1/3

It is known that: - 2 = x (x-1) - (the square of X-Y), ① find the value of X-Y, ② if (x-1) (y + 3) < XY, find the value range of X

X (x-1) - (square of X-Y) = - 2
x²-x-x²+y=-2
∴-x+y=-2
x-y=2
(x-1)(y+3)

Remove the brackets, and then merge the square of the similar term 2x - 0.5 (XY + x) - 8xy

=2x²-0.5xy-0.5x²-8xy
=(2x²-0.5x²)-(0.5xy+8xy)
=1.5x²-8.5xy

If x + y = 5, xy = 4, then the value of xy-2x-2y is

Xy-2x-2y
=4/2-2(x+y)
=2-2*5
=-8

If x, y ∈ R + and xy = 1, then the minimum value of (1 + 1x) (1 + 1y) is ()
A. 4B. 2C. 1D. 14

(1 + 1x) (1 + 1y) = 1 + 1x + 1y + 1xy = 2 + 1x + 1y = 2 + X + YXY = 2 + (x + y) because x, y ∈ R +, and xy = 1, so 2 + (x + y) ≥ 2 + 2XY = 2 + 2 = 4, if and only if x = y = 1, take the equal sign. So the minimum value of (1 + 1x) (1 + 1y) is 4, so select a

Given xy = 3, x + y = 7, try to find the value of (1) x + XY + y, and the value of (2) (X-Y)

1、
(x+y)²=7²
x²+2xy+y²=49
xy=3
Subtract XY from both sides
x²+xy+y²=49-3=46
2、
(x-y)²
=x²-2xy+y²
=(x²+xy+y²)-3xy
=46-3×3
=37

Given x * y + X + y + 1 = 4xy, find the value of XY

x^2*y^2+x^2+y^2+1=4xy
==> x^2*y^2-2xy+1+x^2-2xy+y^2=0
==> (xy-1)^2+(x-y)^2=0
==>Xy = 1 and x = y
==>X = y = 1 or x = y = - 1