Dy divided by DX is equal to the general solution of minus x squared divided by y

Dy divided by DX is equal to the general solution of minus x squared divided by y

dy/dx=-x^2/y
ydy=-x^2dx
∫ydy=-∫x^2dx
1/2y^2=-1/3x^3+c
Namely:
3y^2=-2x^3+C.

If the integral (2x's square + ax-y + 5) - (2bx's square-3x + 5y-1) is simplified without the square term and X term of X, then a = (), B = ()

(2x²＋ax－y＋5)－(2bx²－3x＋5y－1)
=(2－2b)x²＋(a＋3)x－8y＋6
If this formula does not contain terms X and X, then:
2-2b = 0 and a + 3 = 0
A = - 3, B = 1

The second power of X - the square of 4 parts of X + 2x minus 2 parts of X-2
To simplify the process

（x²+2x）/(x²-4)-2/(x-2)
=x(x+2)/(x+2)(x-2)-2/(x-2)
=x/(x-2)-2/(x-2)
=(x-2)/(x-2)
=1

It is proved that the function f (x) = x square + 2x (- 1) - 1 is the square of 2x and is a decreasing function on (0,1),

Its derivative = 2x-2, and the value on (0,1] is less than or equal to 0, so it is a decreasing function in this interval

Find the intersection point P of two lines L: 3x + 4Y-2 = 0: L2 = 2x + y + 2 = 0, and the equation of the line L perpendicular to the line L3: x-2y-1 = 0 is?

3x+4y-2=0
2x+y+2=0
Solving equations
x=-2,y=2
So the intersection (- 2,2)
x-2y-1=0
y=x/2+1/2
Slope 1 / 2
The vertical slope is - 2
So Y-2 = - 2 (x + 2)
So it's 2x + y + 2 = 0

The existence theorem of zero point of continuous function is extended to open interval

If the function f (x) is defined and continuous on the interval (a, b), and there are two different numbers X1 and X2 on (a, b), then f (x1) * f (x2) is satisfied

If mx05-4x05 = x + 1 is a quadratic equation of one variable about X, then the value range of M is

That is, (M-4) x & # 178; = x + 1
If x is a quadratic equation with one variable, then the coefficient of X is not equal to 0
So M-4 ≠ 0
m≠4

As shown in the figure, the image of quadratic function y = 1 / 2x ^ 2-3 / 2x + 1 intersects with y axis at point a, and intersects with X axis at points B and C. m is a moving point on the symmetry axis of function image,
Then the maximum value of | ma-mc | is () a: the difference between radical 5 and radical 2 / 2 B: radical 5 / 2
C: Root 5 d: root 2

Y = (1 / 2) x ^ 2 - (3 / 2) x + 1 = (1 / 2) (X-2) (x-1) a (0,1) B (1,0) C (2,0) axis of symmetry x = (3 / 2) / 2 * (1 / 2) = 3 / 2m (3 / 2, Y0) straight line AC y = (1 / - 2) (X-2) = (- 1 / 2) x + 1 x = 3 / 2, y = (1 / 4) m (3 / 2,1 / 4) | max = | AC | = √ 5My ≠ 1 / 4 | ma-mc|

If two equations 4x = 3 / 2 and 3x + 2m = m-5x have the same solution, then M =?

This is a quadratic equation of two variables
X=3/8 M=-3
It's minus three

Quadratic equation of one variable: Question 1: the 2nd power of 9x - 24x + 14 = 11 question 2: X (3x + 2) (X-7) = 0 question 3: the 4th power of X - the 2nd power of X - 12 = 0
Four questions: 960 (x + 8) = 960x + 20x (x + 8)
To process. Four questions! Give some absolute bonus! To process!

①9x²-24x+14=11
9x²-24x+3=0
∴x1=24+√﹙-24﹚²-4×9×3／2×9=24+6√13／18 x2=24-√﹙-24﹚²-4×9×3／2×9=24-6√13／18
②x﹙3x+2﹚﹙x-7﹚=0
x﹙3x²-21x+2x-14﹚=0
∴x1=0 x2=19+√﹙-19﹚²+4×3×14／2×3=7 x3=19-√﹙-19﹚²-4×3×14／2×3=-2／3
③x∧4-x²-12=0
Let X & # 178; = t  T & # 178; - T-12 = 0
∴t1=1+√1+4×12／2×1=4 t2=1-√1+4×12／2×1=-3
That is, X & # 178; = 4x1 = 2x2 = - 2  X & # 178; ≥ 0  X & # 178; = - 3
④960(x+8)=960x+20x(x+8)
960x+7680=960x+20x²+160x
x²+8x-384=0
∴x1=-8+√8²+4×384／2×1=16 x2=-8-√8²+4×384／2×1=-24