Solution equation: 95-4x = 59

Solution equation: 95-4x = 59

95-4x-59=0
36-4x=0
4x=36
X=9
95-4x=59
4x=36
X=9
What grade are you in?
Solve equation 11x + 5.4 = 17
11X=17-5.4
11X=11.6
X=11.6/11
X=1.05454545454
11X+5.4=17
11x=17-5.4
11x=11.6
x=11.6/11
=58 / 55: how can it be equal to 58 / 55? Isn't 11.6/11 waiting for 1.05454?
How to solve the univariate cubic equation? F (x) = x ^ 3-4x + x-4, f (x) = x ^ 3 + 11x ^ 2 + 39x + 29, f (x) = x ^ 3 + 10x ^ 2 + 33x + 34
So please give me a hand, thanks
Is the value of X when f (x) = 0,
If it is a normal operation, it is usually rational decomposition. For the equation with the first coefficient of 1, if there is a rational root, it must be the whole root, and the root is the factor of the constant term. Generally, we first use the root test method to find an integer root, and then use the long division or split method to decompose the factor, that is, the factor of 1) 4 is ± 1, ± 2, ± 4
1、 The special cubic equation of one variable x ^ 3 + PX + q = 0 (P, Q ∈ R). The discriminant Δ = (Q / 2) ^ 2 + (P / 3) ^ 3. Kardan formula X1 = (Y1) ^ (1 / 3) + (Y2) ^ (1 / 3); x2 = (Y1) ^ (1 / 3) ω + (Y2) ^ (1 / 3) ω ^ 2; X3 = (Y1) ^ (1 / 3) ω ^ 2 + (Y2) ^ (1 / 3) ω, where ω = (- 1 + I3 ^ (1 / 2)) / 2; Y (1,2) = - (q... expansion)
1、 The special cubic equation of one variable x ^ 3 + PX + q = 0 (P, Q ∈ R). The discriminant Δ = (Q / 2) ^ 2 + (P / 3) ^ 3. Kardan formula X1 = (Y1) ^ (1 / 3) + (Y2) ^ (1 / 3); x2 = (Y1) ^ (1 / 3) ω + (Y2) ^ (1 / 3) ω ^ 2; X3 = (Y1) ^ (1 / 3) ω ^ 2 + (Y2) ^ (1 / 3) ω, where ω = (- 1 + I3 ^ (1 / 2)) / 2; Y (1,2) = - (Q / 2) ± ((Q / 2) ^ 2 + (P / 3) ^ 3) ^ (1 / 2). The standard cubic equation AX ^ 3 + BX ^ 2 + CX + D = 0, (a, B, C, D ∈ R, and a ≠ 0). Let x = y-b / (3a) be substituted into the above formula. It can be transformed into a special cubic equation of one variable y ^ 3 + py + q = 0 which is suitable for solving directly by kardan formula. When Δ = (Q / 2) ^ 2 + (P / 3) ^ 3 > 0, the equation has a real root and a pair of conjugate virtual roots; when Δ = (Q / 2) ^ 2 + (P / 3) ^ 3 = 0, the equation has three real roots, one of which has a double root; when Δ = (Q / 2) ^ 2 + (P / 3) ^ 3
Solve the equation 2x & # 178; + 4x = 5
How to solve 2x & # 178; + 4x = 5
A:
2x²+4x=5
2(x²+2x+1-1)=5
2(x+1)²-2=5
2(x+1)²=5+2
(x+1)²=7/2
x+1=±√(7/2)
x=-1±(√14) /2
Divide two sides by two
x²+2x=5/2
x²+2x+1=5/2+1
(x+1)²=7/2
x+1=±√14/2
x=(-2-√14)/2，x=(-2+√14)/2
2x²＋4x＝5
x²+2x=5/2
(x+1)²=5/2+1
(x+1)²=7/2
x+1=±(√14)/2
X = (- 2 + √ 14) / 2 or x = (- 2 - √ 14) / 2
2X & # 178; + 4x = 5 x ^ 2 + 2x = 5 / 2 x ^ 2 + 2x + 1-1 = 5 / 2 (x + 1) ^ 2 = 7 / 2 x + 1 = positive and negative root sign 14 / 2 x = positive and negative root sign 14 / 2 - 1
Set the formula, 2 a minus B minus 4ac under the root sign
When the minimum value of quadratic function y = x2-2x + m is 5, M=______ .
From the minimum value of quadratic function y = x2-2x + m is 5, 4ac − b24a = 4m − 44 = 5, M = 6
Let * be a kind of operation symbol of a certain number. For any real number a and B, a * b = a + 2B / 2. Find the solution of equation 3 * x = 2
3*x=2
(3+2x)/2=2
3+2x=4
2x=1
x=1/2
It's the same as the first floor
Given the fixed point a (a, 0) and the moving point P (x.y) of the ellipse x ^ 2 + 2Y ^ 2 = 8 if 0
Set point P (2 √ 2Sin θ, 2cos θ) PA & sup2; = (2 √ 2Sin θ - a) & sup2; + 4cos & sup2; θ = 4 + 4sin & sup2; θ - 4 √ 2Sin θ a + A & sup2; let sin θ = t, then PA & sup2; = 4T & sup2; - 4 √ 2ta + A & sup2; + 4 minimum is PA & sup2; min = 2A & sup2; - 4A + A & s on T = √ 2A / 2 = sin θ
The minimum value of quadratic function y = x & # 178; - 4x-1 in the range of - 1 ≤ x ≤ 1
y=x²-4x-1
y=x²-4x+4-5
y=(x-2)²-5
-1≤x≤1
When x = 1, y is the smallest
The minimum value is: y = - 4
Please click the [select as satisfactory answer] button below,
Let ⁃ be some kind of operation symbol. For any real number, a ⁃ B = 2a-3b / 3, find the solution of equation (x-1) ⁃ x + 2) = 1
It's easy to understand
Consider (x-1) as a and (x + 1) as B. because a ⁃ B = 2a-3b / 3, so (x-1) ⁃ x + 2) = (2x-2) - (3x + 3) / 3 = 2x-2-x + 1 = X-1, and because (x-1) ⁃ x + 2) = 1, so X-1 is also equal to 1, so x = 2
AB is the two moving points on the ellipse x ^ 2 + 3Y ^ 2 = 1, OA is perpendicular to ob, O is the origin, find the maximum and minimum of ab
Well, that's right. The coordinates of point B on the second floor are wrong. It seems that the coordinates of point a on the third floor are wrong. Let's set the coordinates of point a as (COSA, Sina / √ 3). I find out the coordinates of point B on the second floor (Sina, - √ 3 * COSA). The solution is the same. Finally, we get abmax = 2, abmin = 2 √ 3 / 3
Let the coordinates of point a be (COSA, Sina / √ 3), OA be perpendicular to ob, and angles a and B complement each other, then the coordinates of point B are (Sina, cosa / √ 3)
AB=√((cosa/√3-sina/√3)^2+(sina-cosa)^2)
AB=2√((1-sin2a)/3)
When sin2a = 1, AB has a minimum value of 0, and when sin2a = - 1, AB has a maximum value of 2 √ 6 / 3
The way upstairs is right! But a little bit wrong! Let the coordinates of point a be (sin β, cos β / √ 3) ∵ OA perpendicular to ob by using the parametric equation. Let the coordinates of point B be (sin (β + 90), cos (β + 90) / √ 3), that is, the coordinates of point B are: (COS β, - 1 / √ 3 × sin β) ∵ AB = √ (sin β - cos β) ^ 2 + (1 / √ 3cos β + 1 / √ 3sin β) ^ 2, that is, ab = 2 √ 3 / 3 × √ (1-1 / 2 * sin2 β). Therefore, when sin2 β = - 1, the maximum value of AB is... Expansion
The way upstairs is right! But a little bit wrong! Let the coordinates of point a be (sin β, cos β / √ 3) ∵ OA is perpendicular to ob ∵ and let the coordinates of point B be (sin (β + 90), cos (β + 90) / √ 3), that is, the coordinates of point B are: (COS β, -1 / √ 3 × sin β)! AB = √ (sin β - cos β) ^ 2 + (1 / √ 3cos β + 1 / √ 3sin β) ^ 2, that is, ab = 2 √ 3 / 3 × √ (1-1 / 2 * sin2 β), so when sin2 β = - 1, the maximum value of AB is √ 2, and when sin2 β = 1, the minimum value is 2 √ 3 / 3
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