Using Clem's law to calculate linear equations, the determinant of coefficient is required A. Must not be 0; B. must be 0; C. may be 0; D. may not be 0

Using Clem's law to calculate linear equations, the determinant of coefficient is required A. Must not be 0; B. must be 0; C. may be 0; D. may not be 0

A. {a (1,1) x (1) + a (1,2) x (2)} should be chosen a(1,n)x(n)=b(1)} {a(2,1)x(1)+a(2,2)x(2)… a(2,n)x(n)=b(2)} ………………………………………………………… {a(n,1)x(1)+a(n,2)x(2)… A (n, n) x (n) = B (n)} Cramer's Law: if a linear equation

If we use Cramer's law to solve the system of linear equations a, DET (a) is not equal to 0, then the system of equations has a unique solution, then what is det (a) = 0?

When det (a) = 0, the system of equations may have infinite solutions (for example, each equation is the same), or it may have no solutions (for example, the left side of each equation of the system of equations is the same, but the constant term on the right side is different, that is, there are contradictory equations). It is impossible to judge by using Cramer's rule. This is also the limitation of this theorem

Given that the range of function f (x) = 2x ^ 2 + BX + C / x ^ 2 + 1 (B is less than 0) is [1,3], find the value of real number B, C

Because the definition field of y = f (x) = (2x ^ 2 + BX + C) / (x ^ 2 + 1) is r
So 2x ^ 2 + BX + C = YX ^ 2 + y
(2-y) x ^ 2 + BX + (c-y) = 0
When y = 2, x = (2-C) / b
Because B

If the range of function y = x − BX + 2 on (a, B + 4) (b < - 2) is (2, + ∞), then a + B=______ ．

∵ function y = x − BX + 2 = 1 + − B − 2x + 2 = 1-B + 2x + 2, and ∵ B < - 2, ∵ B + 2 < 0, ∵ function y is a decreasing function on (a, B + 4) (b < - 2), ∵ 4B + 6 < y < a − Ba + 2; and the value range of ∵ y is (2, + ∞), ∵ 4B + 6 = 2, a − Ba + 2 tends to + ∞; ∵ B = - 4, a = - 2, ∵ a + B = (- 4) +

For which values of a, the range of function y = (x + 1) / (x ^ 2 + a) contains [0,1]
Can give a detailed problem-solving process. Thank you~~~

y=(x+1)/(x^2+a)==(x+1)/[(x+1)^2-2(x+1)+a+1]
y [0,1] 1/y>=1
(x+1)-2+(a+1)/(x+1)>=1
Take (x + 1) as X, that is to find the minimum value of the opposite hook function is greater than or equal to 1 + 2
Note that the sign of (a + 1) should be discussed separately
Do it yourself

The range of function f = (x) is [a, b], then the range of y = f (X-2) is [a, b]
As the title
Urgent
With an explanation Be clear, be specific

Let t = X-2, then y = f (X-2) = f (T)
Because the range of y = f (x) is [a, b]
A ≤ f (T) ≤ B, that is, the value range of F (T) is [a, b]
So the range of y = f (X-2) is [a, b]

The range of function y = x + (A / x) (a ≠ 0)

At A0
Using mean inequality
y> = 2 A
So the range is [2 A, + infinity]

(paper) the range of the function y = f (x) defined on R is [a, b], then the range of y = f (x + 1) is ()
A. [a, b] B. [a + 1, B + 1] C. [A-1, B-1] D

∵ the image of the function y = f (x + 1) is obtained by translating the image of the function y = f (x) one unit to the left, and its range of values does not change. The range of values is still [a, b], so a is selected

If the definition field of function f (x + 1) = x2-2x + 1 is [- 2, 0], then the monotone decreasing interval of function y = f (x) is______ ．

The definition field of function f (x + 1) = x2-2x + 1 is [- 2, 0], the equation of symmetry axis of its image is x = 1, then the function y = f (x) is monotonically recursive in the interval [- 2, 0], that is, the decreasing interval is [- 2, 0], so the answer is: [- 2, 0]

1. It is known that 2 | 3A + 4 | + | 4B + 3 | = - | C + 1|
Finding the value of the opposite number of A-B + C

The first thing to remember is that the absolute value must be ≥ 0
Since the left side of the equation must be ≥ 0, so the right side must also be ≥ 0
But the right side must be less than or equal to 0,
So the right side can only be 0, and so can the left side
c=-1,a=-4/3,b=-3/4
a-b+c=-19/12
Opposite number 19 / 12