(A-1) × () = the square of a-2a + 1 emergency

(A-1) × () = the square of a-2a + 1 emergency

Fill in (A-1)

A square (a + 1) + 2A (a + 1) + A + 1

Factorization?
a²（a+1）+2a（a+1）+a+1
=a²（a+1）+2a（a+1）+(a+1)
=(a+1)(a²+2a+1)
=(a+1)(a+1)²
=(a+1)³

2A [B + 1 / 2 [{the square of a - the square of B}] - the square of {a + B}
After decomposition, please bring in a = 3, B = - 2

If a = 3, B = - 2, then
a+b=1 ab=-6
therefore
simple form
=2ab+a(a²-b²)-(a+b)²
=2ab+a(a+b)(a-b)-(a+b)²
=2ab+a(a-b)-1²
=2ab+a²-ab-1
=ab+a²-1
=-6+9-1
=2

If a + B = 1, then 9-A + B = ----, if the square of a + a = 0, then the value of 2A square + 2A + 2008 is-----

1. Please check whether the original copy is correct
2. Because the square of a + a = 0
So: 2A square + 2A + 2008 = 2 (square of a + a) + 2008 = 2008

The value range of a divided by the letter A in (3a ^ 2 + 8a-3) is

a÷(3a^2+8a-3)
＝a/[(3a-1)(a+3)]
3a-1 ≠ 0 and a + 3 ≠ 0
A ≠ 1 / 3 and a ≠ - 3

Given that f (x) = min {x ^ 2 + 2tx + T ^ 2-1, x ^ 2-4x + 3} is an even function (t is a real constant), what is the zero point of y = f (x)?
1,3. - 1, - 3,

This is an interesting question
First, the function f is deformed
f(x) = min {x^2 + 2tx + t^2 - 1, x^2 - 4x + 3}
= min { (x+t)^2 - 1, (x-2)^2 - 1 }
= min { (x+t)^2, (x-2)^2 } - 1
= [min { |x+t|, |x-2| }]^2 - 1,
Now I say if we want f to be an even function (that is, let min {x + T |, | X-2 |} be an even function), we must have t = 2
Min {| x + T |, | X-2 |} and min {| x-t |, | x + 2 |} (the one on the right is obtained by substituting - x)
I can definitely find such an X, so that the result of the former taking small is | x + T | and the result of the latter taking small is | x-t | so that we can find x, so that f (x) is not equal to f (- x), so it is not even function
|X + T | > | X-2 |, solve x > (4-T ^ 2) / (2t + 4) = (2-T) / 2 after both sides square, and then let
|X-t | > | x + 2 |, after the two sides are squared, we find that x < (T ^ 2 - 4) / (2t + 4) = (2 + T) / 2,
Because t > 2, so (2-T) / 2 < x < (2 + T) / 2, there really exists such X that the function is not even
Similarly, you can also prove that if T < 2, you can find x such that the opposite inequality holds and the function is not even
So, t = 2, f (x) = [min {| x + 2 |, | X-2 |}] ^ 2 - 1, zero point is to find
Min {| x + 2 |, | X-2 |} = 1. Since the smaller one on the left is definitely one of them, we may as well:
|X + 2 | = 1 try to solve x = - 3 and - 1, which are smaller than the second one, so these two are roots;
In the same way, you can find two roots 1 and 3. So the root of the answer is four
PS:
If even functions have roots, the number of roots can only be 1, 2, 4, 6, 8;
As for the judgment of T = 2, the proof provided just now, but you can quickly draw a function diagram to get that the images of | X-2 | and | x + T | are respectively symmetrical about x = 2 and x = - t, and taking the small function is to find the curve at the bottom of each segment and finally form a broken line shape. From this, we can see that as long as t is not equal to 2, the two symmetry axes are not symmetrical about the Y axis, The graph can't be symmetrical about the Y axis

On every curve of the image with inverse scale function y = k − 1 x, y decreases with the increase of X, then the value range of K is ()
A. k＞1B. k＞0C. k≥1D. k＜1

According to the meaning of the problem, on each curve of the inverse scale function y = k − 1 x image, y decreases with the increase of X, then k-1 ＞ 0 and K ＞ 1 can be obtained

If the product of five rational numbers is negative, then the negative factor has one

When five rational numbers are multiplied, their product is negative, and there are one, three or five negative factors (odd number)

(4x squared - y cubic - 4xy squared + y) divided by (2xy-1)
=I'm not really good at math

There may be something wrong with your subject

The solution of x ^ 3 + 6x ^ 2 + 11x + 6 requires detailed steps, which is troublesome

x^3+6x^2+11x+6
=(x^3+2x^2)+(4x^2+8x)+(3x+6)
=x^2(x+2)+4x(x+2)+3(x+2)
=(x^2+4x+3)(x+2)
=(x+1)(x+3)(x+2)