Given that the value of algebraic formula (x-a) 2 + B is always positive, then the value of B should be () A. Negative number B. non negative number C. non positive number D. positive number

Given that the value of algebraic formula (x-a) 2 + B is always positive, then the value of B should be () A. Negative number B. non negative number C. non positive number D. positive number

The value of ∵ (x-a) 2 + B is always positive, the minimum value of (x-a) 2 is 0, and the value of ∵ B should be positive

The following statement is wrong ()
A. The value of an algebraic formula is unique. B. the number 0 is an algebraic formula. C. The value of an algebraic formula is not necessarily unique, depending on the value of the letter in the algebraic formula. D. when the algebraic formula n + 1 is used to represent the number of people, n can only take natural numbers

The value of the algebraic expression is not unique, such as: | x | = 1, x = ± 1, so it is wrong

To write an algebraic expression containing the letter A, the algebraic expression must be positive no matter what value a takes

Square of a + 1

When k is a value, the polynomial x2-2xy + ky2 + 3x-5y + 2 can be decomposed into the product of two first-order factors?

In this case, we can make the x-2-2xy + ky2 + 3x-5y-2 (x + 2) (x + 1) (x + 2) (x + 2) (x + 1) (x + my + my + 1) (x + NY + 2), that is, the X-2 + (M + n) XY + mny2 + 3x + 3x + (2m + n) y + 2 = (x + 1) (x + 1) (x + 1) (x + 1) (x + 2) (x + 1) (x + 2) (x + 2) (x + 2) (x + 2) (x + 2) (x + 2) (x + 2) (x + 2) (x + 2) (x + 2) (x + 2) (x + 1) (x + my + my + my + my + my + my + 1) (x + my + my + 1) (x + my + my + my + my + my + my + my + my + my + my + my + my + 1) (x + my + my + my + my + my + my + 1) (x + my + my + & nbsp; & nbsp; & nbsp; & nbsp; ③ When k = - 3, the polynomial x2-2xy + ky2 + 3x-5y + 2 can be decomposed into the product of two first-order factors

When finding the value of M, the polynomial x2-y2 + MX + 5y-6 can be factorized

When m is a value, x ^ 2-y ^ 2-x-5y + M can be decomposed into the product of two linear factors

x^2-y^2-x-5y+m
=(x²-x+1/4)-(y²+5y+25/4)-1/4+25/4+m
=(x-1/2)²-(y+5/2)²+m+6
=(x-1/2+y+5/2)(x-1/2-y-5/2)+m+6
=(x+y-2)(x-y-3)+m+6
When m + 6 = 0, i.e. M = - 6, it can be decomposed into the product of two first-order factors

Factorization of a (a + 1) - B (B-1)
Question 2 ax ay + A ^ 2 + BX by + ab
Question 3 ax BX + CX ay + by CY

a(a+1)-b(b-1)
=a^2+a-b^2+b
=(a+b)(a-b)+(a+b)
=(a+b)(a-b+1)
ax-ay+a^2+bx-by+ab
=(x-y)(a+b)+a(a+b)
=(a+b)(x-y+a)
ax-bx+cx-ay+by-cy
=x(a-b+c)-y(a-b+c)
=(x-y)(a-b+c)

Factorization: (a ^ 2-B ^ 2) x ^ 2 + 4abx - (a ^ 2-B ^ 2)
Urgent!

The original formula = (a + b) (a-b) x & # 178; + [(a + b) & # 178; - (a-b) & # 178;] x + (a + b) (a-b)
=(a+b)(a-b)x²+(a+b)²x-(a-b)²x+(a+b)(a-b)
=(a+b)x[(a-b)x+(a+b)]-(a-b)[(a-b)x+(a+b)]
=[(a-b)x+(a+b)][(a+b)x-(a-b)]
=(ax-bx+a+b)(ax+bx-a+b)

Factorization of AB (x 2 + 1) + X (a 2 + B 2)

ab(x^2+1)+(a^2+b^2)x
=abx^2+(a^2+b^2)x+ab
Cross x ^ 2 and constant
a b
b a
=(bx+a)(ax+b)

4. Factorization: x-25=____ 5. Factorization: a-ab=____ .
6. Factorize the following polynomials. ① m-9n; ② ab-4ab; ③ (5x-3) - 16x; ④ the fourth power of 16x-1

x-25=(x+5)(x-5) a-ab=a(a-b) = a(a+b)(a-b) ① m-9n = (m+3n)(m-3n) ② ab-4ab = ab(b-4) ③ (5x-3)-16x = (5x+4√x-3)(5x-4√x-3) ④ 16x^4-1 = (4x+1)(4x-1) = (4x+1)(2x+1)(2x-1)