Remove the negative bracket (- x + 2-x) - 1 first

Remove the negative bracket (- x + 2-x) - 1 first

You missed the square, right
Original = - X & # 178; + 2x + 3 + 2x & # 178; - 2x-2
=(-x²+2x²)+(2x-2x)+(3-2)
=x²+1

Find the general solution of the derivative of the differential equation x ^ 2Y = (x-1) y

Separation of variables method
x^2dy/dx=(x-1)y
dy/y=(x-1)/x^2dx
dy/y=(1/x-1/x^2)dx
Integral: ln | y | = ln | x | + 1 / x + C1
We get: y = CXE ^ (1 / x)

Finding the first derivative of Y is equal to the general solution of x times y

y'=xy
dy/dx=xy
dy/y=xdx
Integral on both sides: ln | y | = x ^ 2 / 2 + C
That is y = ± e ^ (x ^ 2 / 2 + C)

If 3 + 2I is a root of the equation 3x & # 178; + BX + C = 0, then B = C=

3 + 2I is a root of the equation 3x & # 178; + BX + C = 0
Then: 3-2i is another root of the equation 3x & # 178; + BX + C = 0
b=-18,c=39

For help ~ the equation AX ^ 2 + BX + C = 0 (a ≠ 0) has a non-zero root x1, and the equation - ax ^ 2 + BX + C = 0 has a non-zero root x2
The equation AX ^ 2 + BX + C = 0 (a ≠ 0) has a non-zero root x 1, and the equation - ax ^ 2 + BX + C = 0 has a non-zero root x 2. It is proved that the equation (A / 2) x ^ 2 + BX + C = 0 must have a root between x 1 and x 2

Proof: because ax1 ^ 2 + BX1 + C = 0, so (A / 2) X1 ^ 2 + BX1 + C = - (A / 2) X1 ^ 2
And because - AX2 ^ 2 + bx2 + C = 0, so (A / 2) x2 ^ 2 + bx2 + C = (3a / 2) x2 ^ 2, Let f (x) = ax ^ 2 + BX + C, then: [(A / 2) X1 ^ 2 + BX1 + C] [(A / 2) x2 ^ 2 + bx2 + C] = - (3a ^ 2 / 4) (x1x2) ^ 2 < 0
So the equation (A / 2) x ^ 2 + BX + C = 0 must have a root between x 1 and x 2

Given that the solution of the quadratic equation 2x ^ 2 + BX + C = 0 is X1 = 3, X2 = - 1, then the quadratic trinomial 2x ^ 2 + BX + C can be decomposed into?
There must be a process!

The solution of 2x ^ 2 + BX + C = 0 is X1 = 3, X2 = - 1
-b/2=3-1=2,b=-4
c/2=-3,c=-6
2x^2+bx+c
=2x²-4x-6
=1(x²-2x-3)
=2(x-3)(x+1)

Let f (x) = x ^ 2 + BX + C, two real roots X1 of the equation f (x) = 2x, X2 satisfy x2-x1 > 2
Let f (x) = x ^ 2 + BX + C (B, C are constants), two real roots X1 and X2 of the equation f (x) = 2x satisfy x2-x1 > 2
(1) Verification: B ^ 2 > 4 (B + C);
(2) Let t

Solution 1) Let G (x) = x ^ 2 + BX + c-2x, then X1 and X2 are the two roots of G (x)
x1+x2=2-b,x1x2=c
(x2-x1)^2=(x1+x2)^2-4x1x2=(2-b)^2-4c=b^2+4-4b-4c>4
B ^ 2 > 4 (B + C)
2) According to the sign of C, judge the sign of G (x) and compare f (T) with 2x1

It is known that the parabola y = ax & # 178; + BX + 2 intersects with the x-axis at points a (x1,0), B (x2,0) (x1 < x2) x1, where X2 is two parts of the equation x & # 178; - 2x-3 = 0
Point C is the intersection of the parabola and the y-axis
(1) Finding the value of a and B
(2) Point q is a point on the symmetric axis of parabola. Find the minimum perimeter of △ QAC
(3) If the moving straight line y = m (m < 0) intersects the parabola at d (D is on the left and E is on the right), is there a point P on the x-axis such that △ dep is an isosceles right triangle? If so, the coordinates of point P are obtained; if not, the reason is given

(1) The univariate quadratic equation x ^ 2-2x-3 = 0 (x-3) (x + 1) = 0 is X1 = - 1, X2 = 3 (x1)

If x + ax + B = 0 and X + BX + a = 0 have only one common root, then () a, a = B, B.A + B = 0, C.A + B = 1, D.A + B = - 1

When x ^ 2 + ax + B = 0... (1) x ^ 2 + BX + a = 0... (2) (1) - (2): (a-b) x = (a-b) a = B: two equations have two identical roots, do not take; = = > x = 1, = = = > 1 ^ 2 + A * 1 + B = 0 = = = > A + B = - 1, so D

If a is a root of the equation x2 + BX + 2x = 0 about X, find the value of a + B

Substituting a into: A2 + AB + 2A = 0, that is, a (a + B + 2) = 0
If a ≠ 0, then a + B = - 2
If a = 0, then a + B = B