If a + B + C = 5, the second power of a + the second power of B + the second power of C = 3, find the value of AC + BC + ab

If a + B + C = 5, the second power of a + the second power of B + the second power of C = 3, find the value of AC + BC + ab

(a+b+c)^2=a^2+b^2+c^2+2(ac+ab+bc)=25
ab+bc+ac=11

Given AB = 2, find the value of a 2 − AB + 3B 2 a 2 + AB + 6B 2

From ab = 2, we get a = 2B, then the original formula = 4B2 − 2B2 + 3b24b2 + 2B2 + 6b2 = 512

Let a 2-AB = 1, 4 ab-3 B 2 = - 3. Find the value of a 2-9 AB + 6 B 2-7

∵a2-ab=1，4ab-3b2=-3，∴a2-9ab+6b2-7=（a2-ab）-2（4ab-3b2）-7=1-2×（-3）-7=1+6-7=0．

Given a △ B = 2, find the value of (A & # 178; - AB + 3B & # 178;) / (A & # 178; + AB + 6B & # 178;)

If a > 0, b > 0, and a (a + b) = 3B (a + 5b), find the value of 2A + 3B + aba-b + ab

From the known conditions, a-2ab-15b = 0 is obtained, that is, (a-5b) (a + 3b) = 0, ∵ a > 0, b > 0, ∵ a-5b = 0, the solution is a = 25B, ∵ the original formula = 2 × 25B + 3B + 25b225b-b + 25b2 = 58b29b = 2

The square of a plus the square of AB minus the square of 6B equals zero. Find a + B / b-a

Given 2A = B (AB is not equal to 0), then the square of a -- the square of B, the value of AB is

ab/(a^2-b^2)=2a^2/(a^2-4a^2)=-2/3

(1) Given a & # 178; - AB = 2,4ab-3b & # 179; = - 3, try to find the value of a & # 178; - 13ab + 9b & # 178; - 5
(2) Given M-N = 2, Mn = 1, try to find the value of polynomial (- 2Mn + 2m + 3n) - (3MN + 2n-2m) - (M + 4N + Mn)

a²-ab=2 (1)
4ab-3b³=-3,(2)
(1) (2) × 3
a²-ab-12ab+9b²=2+9
a²-13ab+9b²-5=11-5=6
(2) It is known that M-N = 2, Mn = 1
（-2mn+2m+3n）-（3mn+2n-2m）-（m+4n+mn）
=-2mn+2m+3n-3mn-2n+2m-m-4n-mn
=3m-3n-6mn
=3(m-n)-6mn
=6-6
=0

Given that the quadratic power of a minus AB = 2, the quadratic power of 4AB minus 3B = - 3, find the value of the quadratic power of a minus 13ab plus 9b minus 5

A:
From a ^ 2-AB = 2,4ab-3b ^ 2 = - 3, we can get the following results:
a^2-13ab+9b^2-5
=a^2-ab-12ab+9b^2-5
=(a^2-ab)-3(4ab-3b^2)-5
=2-3*(-3)-5
=2+9-5
=6

If we know a square + AB = 4, AB + b square = - 2, then a square - b square =? A square + 4AB + 3B square =? A square?

Let a square + AB = 4 be Formula 1 and ab + b square be formula 2
So formula 1-2 = a square + ab - (AB + b square) = a square + ab-ab-b square = a square - b square = 4 - (- 2) = 6
So a square - b square = 6
Because a square + 4AB + 3B square = a square + AB + 3 (AB + b square)
=A square + AB + 3AB + 3B square
=A square + 4AB + 3B square
=4+3*（-2）=-2
So a square + 4AB + 3B square = - 2