Given a > b > C, M = A2B + B2C + C2A, n = AB2 + BC2 + Ca2, then the relationship between M and N is () A. M ＜ NB. M ＞ NC. M = nd

Given a > b > C, M = A2B + B2C + C2A, n = AB2 + BC2 + Ca2, then the relationship between M and N is () A. M ＜ NB. M ＞ NC. M = nd

∵ M-N = (A2B + B2C + C2A) - (AB2 + BC2 + Ca2), = A2B + B2C + c2a-ab2-bc2-ca2, = A2 (B-C) + B2 (C-A) + C2 (a-b), = A2 (B-C) + BC (B-C) - AB2 + ac2, = A2 (B-C) + BC (B-C) - A (B + C) (B-C), = (B-C) (A2 + BC AB AC), = (B-C) (A-C) (a-b), and a > b > C, ∵ M-N = (B-C) (A-C) (a-b) > 0, that is, M > n
1 / 2 x square-18 first extract the common factor, then use the square difference formula to decompose the factor? Excuse me?
1 / 2 x squared-18
=1 / 2 (square of x-36) extract 1 / 2 (extract common factor - number that can be extracted)
=1 / 2 (x + 6) (X-6) square difference formula
A 2 + B 2 + C 2-ab-bc-ca = 0 is known, and a = b = C is proved
It is proved that: ∵ A2 + B2 + c2-ab-ac-bc = 0 ∵ 2A2 + 2B2 + 2c2-2ab-2ac-2bc = 0 (a-b) 2 + (B-C) 2 + (A-C) 2 = 0 ∵ A-B = 0, B-C = 0, a-c = 0, ∵ a = B, B = C, C = a, ∵ a = b = C
Which of the following polynomials can be factorized by formula?
Among the following polynomials, the one that can be factorized by formula method is ()
A.-a²-b² B.a²+b² C.-4a²+12ab-9 D.25m²+15m+9
Your C is wrong
It is - 4A & # 178; + 12ab-9b & # 178;
=-(4a²-12ab+9b²)
=-(2a-3b)²
Choose C
No,
Maybe your C is wrong
If C is correct, there is no answer
A. - A & # 178; - B & # 178; = - (A & # 178; + B & # 178;) if this can be decomposed, B can do the same
Then look at C. without the quadratic term of B, it's impossible to decompose
Item D, even if it can be decomposed, is not an acceptable answer: it can be decomposed into 25 (M - (- 15 + (675) ^ 0.5i) / 50) (M - (- 15 - (675) ^ 0.5i) / 50) imaginary numbers... ... unfold
If C is correct, there is no answer
A. - A & # 178; - B & # 178; = - (A & # 178; + B & # 178;) if this can be decomposed, B can do the same
Then look at C. without the quadratic term of B, it's impossible to decompose
Item D, even if it can be decomposed, is not an acceptable answer: it can be decomposed into 25 (M - (- 15 + (675) ^ 0.5i) / 50) (M - (- 15 - (675) ^ 0.5i) / 50) imaginary numbers... Put it away
Among the following polynomials, the one that can decompose the factor by formula is (c)
A. The square of X - the square of XY b.x + the square of XY c.x - the square of y the square of D.X + the square of Y
(2) The following formula is the complete square formula of (a)
A. Square of X - x + quarter B.1 + square of X c.x + XY + 1 square of D.X + 2x-1
July E5
If a + B + C = 4, AB + BC + Ca = 3, then what is the power of a + B + C?
In the range of rational numbers, the following polynomials can be factorized by formula method
A. The square of A-6A
B. The square of a - the square of AB + B
C. The square of a - AB + the square of a quarter of B
D. Square of a - quarter of the square of AB + B
What does "formula method" mean
Option C:
a^2 - ab - (b^2)/4
=a^2 - ab/2 - ab/2- (b^2)/4
=a(a - b/2) -(b/2)(a - b/2)
=(a - b/2)*(a - b/2)
=(a- b/2)^2
Reverse deduction:
(a- b/2)^2
=a^2 - 2*a*(b/2) +(b/2)^2
=The square of a - AB + the square of a quarter of B
Simplify the square of (a-5) (a + 1) / a-5a divided by (A's square + a)
(a-5)(a+1)/a-5a/(a²+a)=(a-5)(a+1)/a-5/(a+1)=[(a-5)(a+1)²-5]/[a(a+1)]=[(a-5)(a²+2a+1)-5)]/[a(a+1)]=(a³+2a²+a-5a²-10a-5)/[a(a+1)]=(a³-3a²-9a-10)/[a(a+1)]
-1 / 2x ^ 2 + 2XY XZ factorization
-1/2x^2+2xy-xz
=x(-x/2+2y-z)
(5a's square-2a) - (A's square-5a + 1) reduction
(square of 5a-2a) - (square of a-5a + 1)
=5a²-2a-a²+5a-1
=4a²+3a-1
(square of 5a-2a) - (square of a-5a + 1)
=5a^2-2a-a^2+5a-1
=4a^2+3a-1
The square of 3A + 4A - 1
Factorization 8A ^ 3B ^ 2-12ab ^ 3C + 4AB - 1 / 2x ^ 2 + 2XY XZ 2 ^ 2011-2 ^ 2010
-4x^4+2x^3y [9x+3)^2+(x+3)(x-3)]/2x
8a^3b^2-12ab^3c+4ab
=4ab(2a²b-3b²c+1)
-1/2x^2+2xy-xz
=-1/2x(x+4y-2z)
2^2011-2^2010
=2^2010x2-2^2010
=2^2010(2-1)
=2^2010
-4x^4+2x^3y
=-2x^3(2x-y)
[9x+3)^2+(x+3)(x-3)]/2x
=81x^2+9+54x+x^2-9/2x
=82x^2+54x/2x
=41x+27
8a^3b^2-12ab^3c+4ab
=4ab(2a²b-3b²c+1)
-1/2x^2+2xy-xz
=-x(x-4y+2z)/2
2^2011-2^2010
=2^2010(2-1)
=2^2010
-4x^4+2x^3y
=-2x³(2x-y)
[9x+3)^2+(x+3)(x-3)]/2x
=(81x²+54x+9+x²-9)/2x
=41x+27
... speechless