Ax ^ 2 + 4x + 4 ≥ - 2x ^ 2 + 1 for X ≤ R When a = 0 When a ≠ 0

Ax ^ 2 + 4x + 4 ≥ - 2x ^ 2 + 1 for X ≤ R When a = 0 When a ≠ 0

(a+2)x²+4x+3≥0
a=-2
Then 4x + 3 ≥ 0 is not constant
a≠-2
Then on the left is a quadratic function
Constant greater than or equal to 0
So opening up, a + 2 > 0, a > - 2
And the minimum value is greater than or equal to 0
So the function and X cycle have no intersection or have an intersection, so the discriminant is less than or equal to 0
So 16-12 (a + 2) ≤ 0
a≥-2/3
Comply with a > - 2
So a ≥ - 2 / 3
The minimum value of polynomial x-y-2x + 4Y + 7
formula
Original formula = = x ^ 2-2x + 1-y ^ 2 + 4y-4 + 10 = = (x-1) ^ 2 - (Y-2) ^ 2 + 10
Because - (Y-2) ^ 2 can be infinitely small, the original formula has no minimum value
You must have copied the wrong question! Maybe it's x ^ 2-2x + y ^ 2 + 4Y + 7. In that case, the minimum value is 2
(1) 5-3x of 2 = 3-5X of 3 (2) 1-2x of 1 = 3-6 (3) y of 4 + 2y-1 of 2-6 = 1
1.15-9x=6-10x
x= -9
2.x= -6
3.3y-4y= -14
y=14
X ^ 4-2x & # 178; = 0 is solved by substitution method
Let a = x & # 178;
Then x ^ 4 = A & # 178;
a²-2a=0
a(a-2)=0
a=0,a=2
x²=0,x²=2
So x = 0, x = - √ 2, x = √ 2
Let t = x & # 178;
t²-2t=0
t1=0
t2=2
x²=0
x1=0
x²=2
X2 = root 2
X3 = - radical 2
X1 = 0; x2 = radical 2; X3 = - radical 2
Let X & # 178; = m, then the original formula is equal to M & # 178; - 2m = 0
M = 0 or M = 2
That is X & # 178; = 0 or X & # 178; = 2
X=0
X = root 2
X = negative root 2
Let x ＾ 2 = t (T > 0), then t ＾ 2-2t = 0, the solution is t = 0 or 2
That is, X ＾ 2 = 0 or 2
The solution is x = 0 or x = positive and negative root sign 2
It is known that the parabola y = ax ^ 2 + BX + C passes through the point (- 1,1), and for any real number x, 4x-4 is less than or equal to AX ^ 2 + BX + C is less than or equal to 2x ^ 2-4x + 4
(1) Find the value of 4A + 2B + C (2); find the analytic formula of y = ax ^ 2 + BX + C
(1) Let x = 2, then 4 ≤ 4A + 2B + C ≤ 4, ｜ 4A + 2B + C = 4; (2) ∵ the parabola passes (- 1,1), ｜ A-B + C = 1, ｜ B = 1-A, C = 2-2a, and ax ^ 2 + BX + C ≥ 4x-4, ax ^ 2 - (a + 3) x + 6-2a ≥ 0, ax ^ 2 - (a + 3) 2-4a (6-2a) ≥ 0, that is, (A-1) 2 ≤ 0, ｜ a = 1, and when a = 1, X
When x and y are the values, the polynomial X & # 178; + 4Y & # 178; - 8x + 12Y + 5 has the minimum value
simple form
=(x²-8x+16)+(4y²+12y+9)-20
=(x-4)²+(2y+3)²-20
The least square is 0
So x-4 = 0, 2Y + 3 = 0
That is, when x = 4 and y = - 3 / 2, the minimum value is - 20
The solution equations are: 1, 91 △ x = 1.32, 410-3x = 1703, 0.5x + 8 = 434, 8x △ 1.6 = 35, 4x-3.2 × 30 = 24
X = unknown, if you can solve other equations, please reply!
1、91÷x=1.3 2、410-3x=170 3、0.5x＋8=43 4、8x÷1.6=3 5、4x-3.2×30=24
x=91÷1.3 3x=410-170 0.5x=43-8 8x=3x1.6 4x-96=24
x=70 3x=240 0.5x=35 8x=4.8 4x=24+96
x=80 x=70 x=0.6 4x=120
x=30
X=70
X=80
X=70
x=0.6
x=30
1、91÷x=1.3
91 △ x times 91 = 1.3 times 91
X = 1183 sorry, no time
1+1+1+1+1+9+9+9+9
F (2x + 1) = x & # 178; + 1 to find f (x) with the substitution method, the steps should be detailed
I really don't understand the step of solving equations, although the problem is very simple Please wait online
Let 2x + 1 = t, then x = (t-1) / 2
f(t)=[(t-1)/2]²+1=(t²-2t+5)/4=t²/4-t/2+5/4
T is also a value in the domain, and t is replaced by X
f(x)=x²/4-x/2+5/4
If the square of ax plus 4x plus a is greater than or equal to minus 2x plus 1, the range of a can be obtained
Constant ax & # 178; + 4x + a ≥ - 2x & # 178; + 1
(a+2)x²+4x+(a-1)≧0.
And ⊿ = 16-4 (a + 2) (A-1) ≤ 0
∴a≧2
Does the polynomial X & # 178; - 2x + 2 have a minimum? If so, what is the value of X when it has a minimum?
x²-2x+2
=(x-1)²+1
The minimum value is 1, where x = 1