Let X and y be nonnegative real numbers, and x ^ 2 + y ^ 2 = 4, u = XY - 4 (x + y) + 10, find the maximum value of U

Let X and y be nonnegative real numbers, and x ^ 2 + y ^ 2 = 4, u = XY - 4 (x + y) + 10, find the maximum value of U

Let x = 2cosa, y = 2sina, then u = 4cosasina-4 (2sina + 2cosa) + 10 = 2 [(Sina + COSA) & sup2; - 1] - 8 (Sina + COSA) + 10 = 2 [(Sina + cosa-2) & sup2; because - √ 2 ≤ Sina + Cosa ≤ √ 2, then UMIN = 2 (√ 2-2) & sup2; = 4 (3-2 √ 2) Umax = 2 (- √ 2-2) & sup2; = 4 (3 + 2 √ 2)

Sin (x ^ 2 + y ^ 2) + e ^ x-xy ^ 2 = 0

sin (x^2+y^2)+e^x-xy^2=0
Left and right differential
cos(x^2+y^2)*(2xdx+2ydy)+(e^x)dx-(y^2)dx-2xydy=0
The rest is just dy

X ^ 2-y ^ 2 = XY, find X / y,

Divide y by Y on both sides;
(x/y)²-1=x/y
(x/y)²-(x/y)-1=0
From the root formula, we can get X / y = (1 + √ 5) / 2 or (1 - √ 5) / 2

x+y=-3 xy+x+y=-2 x/y+y/x=?

x+y=-3
xy+x+y = -2
Subtract the two formulas to get xy = 1
x/y + y/x
= (x^2+y^2) / xy
= ( (x+y)^2-2xy) / xy
= (9-2)/1
= 7

If 1 ≤ x2 + Y2 ≤ 2, then the value range of x2 + XY + Y2 is given______ ．

Let x = asin θ, y = ACOS θ, t = x2 + XY + Y2, then 1 ≤ a ≤ 2 can be obtained if 1 ≤ x2 + Y2 ≤ 2, then t = x2 + XY + y2 = A2 + a2sin θ cos θ = (1 + 12sin2 θ) A2, from the properties of trigonometric function, 12 ≤ (1 + 12sin2 θ) ≤ 32 can be obtained, so 12 ≤ t ≤ 3, so the answer is [12, 3]

If x > y and xy = 1, then the value range of x ^ 2 + y ^ 2 / X-Y is given
The answer is [2 radical 2, positive infinity]

If AB = C (C is a constant), then a + B ≥ 2 √ (AB) if the above can be understood: --- - --- (x ^ 2 + y ^ 2) / (X-Y) = (x ^ 2 + y ^ 2-2 + 2) / (X-Y) = (x ^ 2 + y ^ 2-2xy + 2) / (x -...)

Let x > 0, Y > 0, and XY - (x + y) = 1, find the value range of X + y

XY ≤ [(x + y) / 2] ^ 2, so xy = 1 + X + y ≤ [(x + y) / 2] ^ 2, that is, (x + y) ^ 2-4 (x + y) - 4 ≥ 0, the solution is x + y ≤ - 2 √ 2 + 2 (rounding) or x + y ≥ 2 √ 2 + 2
x+y≥2√2+2

X = 2A + 1, y = A-2, if XY ＞ 0, find the value range of A

X = 2A + 1, y = A-2, if XY ＞ 0, find the value range of A
Because x = 2A + 1, y = A-2, XY > 0
So (2a + 1) (A-2) > 0
When 2A + 1 > 0, A-2 > 0 or 2A + 10
The solution is: a > - 1 / 2, a > 2, so a > 2
When 2A + 1

If x + y = XY, then the value range of X + y is given

Let x + y = t, y = t-x substitute x + y = XY
t=x(t-x)
x²-tx+t=0
Δ=t²-4t≥0
‖ t ≤ 0, or t ≥ 4
The value range of X + y is
（-∞,0]U[4,+∞)

Given x > 0, Y > 0, if xy = x + y + 1, what is the value range of X + y

It's relatively simple ~ the average is OK~
x+y=xy-1=0
So according to the root formula, we can get x + Y > = 2 √ 2 + 2 or x + y = 2 √ 2 + 2