We know the inequality system 1 ≤ KX & sup2; + 2x + K ≤ 2 for X It's not finished. Wrong press. If there is a unique real solution, then the value set of the real number k is? That's KX ^ 2

We know the inequality system 1 ≤ KX & sup2; + 2x + K ≤ 2 for X It's not finished. Wrong press. If there is a unique real solution, then the value set of the real number k is? That's KX ^ 2

1 ≤ KX ^ 2 + 2x + K ≤ 2 has unique real solution
When k > 0, KX ^ 2 + 2x + k = 2
So 4-4k (K-2) ≥ 0
k^2-2k-1≤0
- radical 2 + 1 ≤ K ≤ (radical 2) + 1
So 0
It is known that the solution of the inequality 2x + 5 ≤ 3 (x + 2) is a part of the solution of the inequality (x + 5) / 2 + 1 > (KX + 2) / 2, and the value range of K is obtained
RT
Why 1-k
2x+5≤3(x+2)
x>=-1
（x+5）/2 +1>（kx+2）/2
x+5+2>kx+2
(1-k)x>-5,1-k0,k5/(k-1)
The solution x > = - 1 of 2x + 5 ≤ 3 (x + 2) is a part of the solution x > 5 / (k-1) of (x + 5) / 2 + 1 > (KX + 2) / 2
5/(k-1)=-4
Or - 4
Left solution set [1, ∞)
The solution set on the right is reduced to (k-1) x0 inequality by dividing both sides of the inequality by k-1 (invariant inequality sign) at the same time (- ∞), 5 / (k-1) must not include the previous solution set, so it does not conform to the requirement
3:k
The solution of inequality about X: 2x ^ 2 + kx-k ≤ 0 (k is all real numbers)
1) When △ = k ^ 2 + 8K
Calculation: ① 3x & # 178; · 4x=____ ②- 4/5a³·（- 2/5a）²=___
③(-3mn²)·(2m²n)=___ ④(2.5×10²)×(4×10³)=______ ,3a²b·2abc· 1/3 abc²=____ ⑤1/2 xy²·6x²y
① The third power of 12x, the fifth power of 15A - 125th, the third power of N - 6m, the fifth power of 10, the fourth power of 2a, the third power of B, the third power of C, the third power of 3x, the third power of Y,
3x-2y = 1,5x + 3Z = 8,3y-6z = - 1
fast
3x-2y=1 (1)
5x+3z=8 (2)
3y-6z=-1 (3)
(2)×2+(3)
10x+3y=15 (4)
(1)×3+(4)×2
9x+20x=3+30
29x=33
therefore
x=33/29
y=(3x-1)/2=35/29
z=(3y+1)/6=67/87
X=33/29 Y=35/29 Z=67/29
Given that the image of function f (x) = loga (2 + ax) and the image of function g (x) = log1a (a + 2x) (a > 0, a ≠ 1) are symmetric with respect to the line y = B (B is a constant), then a + B=______ .
g(x)＝log1a(a+2x)=-loga（a+2x） It is known that if M (x, y) is any point in the f (x) image, then the symmetric point m '(x, 2b-y) of M symmetric with respect to the line y = B must be on the image of G (x). Substituting the coordinates of two points into the corresponding analytic expressions, we can get y = loga (2 + ax), 2b-y = - loga (a + 2x). Adding the two expressions, we can get 2B = loga (2 + ax) - loga (a + 2x) = loga2 + AXA + 2x, so 2 + AXA + 2x = 1 The solution is a = 2, so B = 0, so a + B = 2, so the answer is: 2
X²-4X+Y²+6Y+13=0 X=?Y=?
X & sup2; - 4x + Y & sup2; + 6y + 13 = 0 x =? Y =? Fast
13=4+9
So (X & sup2; - 4x + 4) + (Y & sup2; + 6y + 9) = 0
(X-2)²+(Y+3)²=0
The square is greater than or equal to 0, and the sum is equal to 0
If one is greater than 0, then the other is less than 0
So both are equal to zero
So X-2 = 0, y + 3 = 0
X=2,Y=-3
X²-4X+Y²+6Y+13=0
(x-2)^2+(y+3)^2=0
x=2,y=-3
(x-2)^2+(y+3)^2=0
x-2=0,x=2
y+3=0,y=-3
Circular area formula. Two kinds
The circumference of a circle C = 2 π r = or C = π D
The area of circle s = π R ^ 2;
Sector arc length L = n π R / 180
Sector area s = n π R ^ 2 / 360 = LR / 2 (L is the arc length of sector)
The diameter of the circle d = 2R
S=πr^2
S=π(d/2)^2
S=π(C/2π)^2
πR²
S = R & # 178; π--- R radius
S = D & # 178; π / 4 ------ D diameter
Simplify the following formulas: (1) 8x - (- 3x-5) (2) (3x-1) - (2-5x) (3) (- 4Y + 3) - (- 5y-2) (4) 3x + 1-2 (4-x)
(1) Original formula = 8x + 3x + 5
=11x+5
(2) Original formula = 3x-1-2 + 5x
=8x-3
(3) The original formula = - 4Y + 3 + 5Y + 2
=5+y
(4) the original formula is 3x + 1-8 + 2x
=5x-7
(1)8x+3x+5=11x+5
(2)3x-1+2+5x=8X+1
(3)-4y+3+5y+2=y+5
(4)3x+1-8+2x=5x-7
（1）=8x+3x+5=11x+5
(2)=3x-1-2+5x=8x-3
(3)=-4y+3+5y+2=y+5
(4)=3x+1-8+2x=5x-7
Hello LZ, it takes patience and care to simplify these equations. Master the four basic algorithms and merge the similar terms.
（1）8x-（-3x-5） =8x+3x+5=11x+5
（2）（3x-1）-（2-5x） =3x-1-2+5x=8x-3
（3）（-4y+3）-（-5y-2）= （-4y）+3+5y+2）= y+5
（4）3x+1-2（4-x）=3x+1-8+2x=5x-7
Methods: the results were summarized as follows
The following equations are solved by substitution elimination method: 2x + 3Y = 49,3x-2y = 15
3x-2y=15
So y = (3x-15) / 2
Substituting 2x + 3Y = 49
2x+(9x-45)/2=49
2x+9x/2-45/2=49
13x/2=143/2
So x = 11
y=(3x-15)/2=9
2x+3y=49 x=（49-3y）/2
Substituting x into 3x-2y = 15
3/2（49-3y）-2y=15
147-9y-4y=30
Y=9
x=11
2x+3y=49, （1）
3x-2y=15 (2)
(1) X 2 + (2) × 3
13x=143
x=11
Substituting (1) to get:
3y=27
Y=9
Therefore, the solution of the equations is: x = 11; y = 9
The answer is very good~~~
wuyu