If the intersection points of parabola y = 2x ^ 2-4x + 1 and X axis are (x1,0), (x2,0), then x! ^ 2 + x2 ^ 2=

If the intersection points of parabola y = 2x ^ 2-4x + 1 and X axis are (x1,0), (x2,0), then x! ^ 2 + x2 ^ 2=

In the analytic formula y = 2x ^ 2 - 4x + 1, a = 2, B = - 4, C = 1
According to Weida's theorem, we can get X1 + x2 = - B / a = 2; X1 * x2 = C / a = 1 / 2
x1^2 + x2^2
= ( x1 + x2)^2 - 2 * x1*x2
= 4 - 2* (1/2)
=4 - 1
=3

How many arbitrary constants are included in the general solution of the differential equation y '' '+ (e ^ x) y' '+ e ^ (2x) = 1?

The first is the linear equation
And then we'll see how many arbitrary constants there are at the highest derivative
Because the solution of n-order ordinary differential equation is just like the integration of n-times, each integration will have an arbitrary constant
therefore
Here the highest derivative is of order 3 (y '')
So there are three arbitrary constants

General solution of differential equation (y ^ 3 + XY) y '= 1

(y^3+xy)y'=1
X '= XY + y ^ 3 is a first-order linear equation with X as an unknown function
x=e^(y^/2)(C+∫y^3e^(-y^/2)dy)
=e^(y^/2)(C-(y^2+2)e^(-y^/2))
=Ce^(y^/2)-(y^2+2)

If f (2 / x + 1) = lgx, then f (x)=
This is a problem, is to find the analytic function

Let 2 / x + 1 = t
The solution is t = 2 / (t-1)
So f (T) = LG [2 / (t-1)] (T > 1)
So f (x) = LG [2 / (x-1)] (x > 1)

A room 5.2 meters long, 3 meters wide and 6 meters wide, the area of doors and windows is 12 m2. The four walls and roof should be painted, and the painting surface should be smooth
For a room 5.2 meters long, 3 meters wide and 6 meters wide, the area of doors and windows is 12 meters. To paint the four walls and the roof, how many square meters is the painting area? If 200 grams of paint is used per square meter, how many kilograms of paint are needed?

1. The area to be painted is:
5.2×2.6×2﹢3×2.6×2﹢5.2×3﹣12
＝27.04﹢15.6﹢15.6﹣12
＝58.24﹣12
＝46.24﹙㎡﹚
2. The number of coatings required is:
0.2㎏×46.24＝9.248㎏.
A: a total of 9.248 kg of paint is required

There are two points on the line AB, m, N, am: MB = 4:11, n is the midpoint of Ma, and Mn = 1, find the length of AB a --- n --- m -------- B

Because n is the midpoint of AM and Mn = 1, am = 2, and because am: MB = 4:11, MB = 5.5, ab = 2 + 5.5 = 7.5

Two triangles of equal area can form a parallelogram______ (judge right or wrong)

For example: two triangles with the length of 4 and the height of 3 and the length of 2 and the height of 6 have the same area, but they cannot be combined into parallelogram. Two triangles with the same area can be combined into parallelogram, which is wrong

What are PM and am? What do PM and am stand for?

PM: it means afternoon
Am: morning

Calculation: (1) (− 13) − 2 + (136) 0 + (− 5) 3 △ (− 5) 2; (2) (M + 3n) (m-2n) - (2m-n) 2

(1) The former formula is 9 + 1-5 = 5; (2) the former formula is m2-2mn + 3mn-6n2-4m2 + 4mn-n2 = - 3M2 + 5mn-7n2

If a tangent is drawn from a point on the straight line X-Y + 3 = 0 to the circle (x + 2) 2 + (y + 2) 2 = 1, the minimum length of the tangent is ()
A. 322B. 142C. 324D. 322-1

(x + 2) 2 + (y + 2) 2 (x + 2) 2 + (y + 2) 2 = 1 center a (- 2, - 2), straight line X-Y + 3 = 0 any point P, tangent PQ (q is the tangent point), then PQ | (PQ is the tangent point), then PQ | = Pap | 2 − 1, if and only if | PA | PA | is the minimum, it is easy to see the minimum value of | PA | the minimum value, that is the distance from a to the line X-Y + 3 + 3 = 0 X-Y + 3 = 0, the distance D, D, D, D, d = \\124\\\124\124\124it's not easy