How to solve X4 + x2-2ax + 1-a2

How to solve X4 + x2-2ax + 1-a2

X^4+X^2-2aX+1-a^2
=-(a^2+2aX+X^2) + 1+X^4+2X^2-2X^2
=-(a+X)^2+2X^2 + (1+X^2)^2-2X^2
Note: x ^ 2 = the square of X, x ^ 4 = the fourth power of X
A ^ 2 = the square of a, and so on

Factorization of x ^ 2 - (y ^ 2-4y + 4)

x^2-(y^2-4y+4)
=x^2-(y-2)^2
=(x+y-2)(x-y+2)
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Add three plus signs, several minus signs or minus signs between 123456789 to make the result 100

123-4-5-6 + (- 7) + 8 + (- 9) = 100. If you want a minus sign, you can write it like this. In fact, both of them are similar
123-（4+5+6)-7+8-9=100

When 1 / 3A

1/3a

On mathematical problems of rational numbers
Fill - 1, + 2, - 3, + 4, - 5, + 6, - 7, + 8 and - 9 in the square so that the three numbers on each diagonal of each row and column meet the following requirements:
(1) The product of three numbers is negative;
(2) The third uncle is absolutely worth and equal
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Third order magic square: Method 1: 1: 2 3 4 1 2 4 9 24 5 6 → 7 5 3 → 3: 5 7 7 7 8 9 8 9 6 8 1 6 1, fill in 1,2,3,4,5,6,7,8,9,2 from top to bottom, from left to right, 8 numbers of outer box, rotate one lattice clockwise, 3, exchange cross numbers except four corners and center, namely 1 and 9,7 and 3, exchange position

Some calculations related to the square of B minus 4ac of quadratic equation of two variables
1 satisfy the square of (x-3) + the square of (Y-3) = 6 of all real number pairs (x, y), the maximum value of Y divided by X in (x, y) 2 know the square of (x-z) - 4 (X-Y) (Y-Z) = 0 find (x + z) divided by y equal to?

1) Let y = KX
（kx-3)^2+(x-3)^2=6
(k^2+1)x^2-6(k+1)x=12
If the equation has real root condition 36 (K + 1) ^ 2-48 (k ^ 2 + 1) > 0, the solution is 3 - √ 2

Can coefficient a of linear regression equation be negative

Yes

Using definite integral to calculate area in high school mathematics
The area enclosed by parabola y & # 178; = 4x and straight line y = x-3 is

The answers and pictures are good, but the calculation method is too cumbersome
Let's look at this problem from another angle. Since we can use X as the integral variable, we can also use y as the integral variable. The problem is much simpler
Firstly, the two analytic expressions are expressed as x = y & # / 4 and x = y + 3
So the enclosed area is: S = ∫ (y + 3-y & # 178 / 4) dy (integral upper limit is 6, lower limit is - 2) = y & # 178 / 2 + 3y-y & # 179 / 12 (integral upper limit is 6, lower limit is - 2) = 18 + 18-18 - (2-6 + 8 / 12) = 64 / 3

Thinking questions on page 48 of sixth grade mathematics exercise book of Jiangsu Education Press

Which book is it, the character?

It is known that if the ratio of x 2 to 9 plus y 2 to 8 is equal to 1, f is the right focus, P (1,1) is a point in the ellipse, and M is a point on the ellipse, then the minimum value of | MP | plus 3 | MF | is
X 2 is the square of X, Y. similarly, x 2 is the square of X divided by 9,

Let F1 be the left focus of the ellipse, 3 | MF | = 6a-3 | MF1 |, so | MP | + 3 | MF | = | MP | + 6a-3 | MF1 |, if you want to be the minimum, you need to have the minimum value of | MP | - 3 | MF1 |, when m is on the extension line of f1p, you can bring it in and you can solve it. You can calculate it yourself, and the amount of calculation is a bit large