Solve the equation (x-a) ^ 2 = B (x ^ 2-A ^ 2) about X (where A.B is a constant and B is not equal to 1) I'm from the third grade of junior high school. Answer with the knowledge of the third grade of junior high school

Solve the equation (x-a) ^ 2 = B (x ^ 2-A ^ 2) about X (where A.B is a constant and B is not equal to 1) I'm from the third grade of junior high school. Answer with the knowledge of the third grade of junior high school

(x-a)²=b(x²-a²)
(x-a)(x-a)=b(x-a)(x+a)
If x-a = 0, x = a, the equation holds, so x = a is the solution of the equation
If x-a ≠ 0, then both sides of the equation are divided by x-a at the same time
x-a=b(x+a)
x-a=bx+ab
x-bx=a+ab
(1-b)x=a(1+b)
If B ≠ 1, then x = a (1 + b) / (1-B)
Comprehensive available
When B ≠ 1, x = a or x = a (1 + b) / (1-B)

How to solve the equation that x + 4 / x equals 0.22?

X/(X+4)=0.22
X=0.22X+0.88
0.78X=0.88
X=0.88/0.78=44/39

Let me ask you an equation. X * 30% = 70000 + X, how much is x?

X*30%=70000+X
X*0.3-X=70000
-0.7X=70000
x=-100000

We know that P is not equal to 0, and solve the equation about X: 2pq - (P + Q) x = (P-Q) X
We also know that P is not equal to 0, and we solve the equation about X: 2pq - (P + Q) x = (P-Q) X

（p-q+p+q）x=2pq,2px=2pq,x=q

The number of solutions to the equation a ^ x + x ^ 2 = 2 (a > 0, a is not equal to 1) should be explained in detail?
The number of solutions to the equation a ^ x + x ^ 2 = 2 (a > 0, a is not equal to 1) should be explained in detail
I draw and analyze, but I draw four intersections

You can analyze, that's right~
However, a is a constant, it can only belong to one of (0,1) and (1, + ∞), and there are only two solutions in any interval

(a + 2) x - 6 = 2x - 3 (a is not equal to 0)

(a+2)x-6=2x-3
ax+2x-6=2x-3
ax=3
x=3/a

How many solutions are there for equation (a ^ x) + 1 = -- (x ^ 2) + 2x + 2A (a > 0, and a is not equal to 1)

The original equation is a ^ x = - x ^ 2 + 2x + 2a-1
Let f (x) = a ^ x, G (x) = - x ^ 2 + 2x + 2a-1 = - (x-1) ^ 2 + 2A
The number of solutions of the equation is the number of intersections of F (x) and G (x)
If f (x) passes through point (1, a), the vertex of G (x) is (1, 2a), and if a > 0, it is obvious that point (1, a) is inside the image of quadratic function g (x). It can be seen that f (x) and G (x) must intersect at two points, no matter 0

The ellipse x2 / 2 + y2 = 1 and the line with slope 1 intersect at two points a ` B. f is the left focus of the ellipse. Find the maximum area of the triangle ABF?

It is known that the line AB: 3x + 4Y + 12 = 0, the line BC: 4x-3y + 16 = 0, the line CA: 2x + Y-2 = 0 where the three sides of the triangle ABC are
Find: (1) the equation of the straight line where the bisector of angle ABC lies
(2) The equation of the line of the median line parallel to the BC side

(1) The simultaneous line AB: 3x + 4Y + 12 = 0, the line BC: 4x-3y + 16 = 0,
The solution is: point B coordinate (- 4,0);
Let the bisector of angle ABC intersect the straight line Ca at point D, coordinate (x, 2-2x),
The distances from point d to line AB and line BC are equal
|4-x | = | 2x + 2 |, the solution is: x = 2 / 3, or x = - 6 (rounding off),
So the coordinates of point d (2 / 3,2 / 3) are,
So the equation of the line where the bisector of angle ABC is located is as follows:
x-7y+4=0.
(2) The results show that the simultaneous line AB: 3x + 4Y + 12 = 0, the line AC: 2x + Y-2 = 0,
The solution is: point a coordinates (4, - 6);
So the midpoint of AB is (0, - 3),
The slope of the line BC is - 4 / 3, so
The equation of the straight line of the median line parallel to the BC side is as follows:
4x+3y+9=0.

In the parabola, y = ax ^ 2 + BX + C satisfies four conditions: 1. ABC = 0.2. A + B + C = 3.3. AB + BC + Ca = - 4.a

A ≠ 0, otherwise it is not a parabola. ABC = 0
①b＝0.a+c＝3.ac＝-4.a＜b＜c.（c-a）²＝（c+a）-4ac＝9+16＝25.
c-a＝5.∴a＝-1,c＝4.
The parabola is y = - X & sup2; + 4
② C = 0.a ＜ B ＜ C.A ＜ 0, B ＜ 0, ab ＞ 0, which is contradictory to ab = - 4
In a word, only y = - X & sup2; + 4