If two unequal real numbers m, n satisfy m * 2-2m = a, n * 2-2n = a, m * 2 + n * 2 = 5, then the value of real number a is?

If two unequal real numbers m, n satisfy m * 2-2m = a, n * 2-2n = a, m * 2 + n * 2 = 5, then the value of real number a is?

Weida theorem, 1 / 2

If the absolute value of real number 2m + 2 + (n-2) ^ 2 = 0, then m * n =?

The absolute value of 2m + 2 + (n-2) ^ 2 = 0
2m+2=0
m=-1
n-2=0
n=2
mxn=-1x2=-2

Given that 3 times (square of M) - 2m-5 = 0, 5 times (square of n) + 2n-3 = 0, where m and N are real numbers, find the absolute value of M-1 / n

Two equations can be solved. Have you learned this?
3m2-2m-5 = (M + 1) (3m-5) = 0, M = - 1 or 5 / 3
5n2 + 2n-3 = (n + 1) (5n-3) = 0, the solution is n = - 1 or 3 / 5
So M-1 / N = - 1 + 1 = 0, or 5 / 3 + 1 = 8 / 3, or - 1-5 / 3 = - 8 / 3, or 5 / 3-5 / 3 = 0
So the answer is 0 or 8 / 3

1. Simplify first, then evaluate 1 / 3A (3a ^ 3-6a ^ 2 + 3a), where a = 7 2. Divide X / 2Y and 2 / 3xy ^ 2 (urgent!

1.
1/3a(3a^3-6a^2+3a）
=a^2-2a+1
=49-14+1
=36
two
X / 2Y and 2 / 3xy ^ 2
x/2y=3x/6y=3x^2y/6xy^2
2/3xy^2=4/6xy^2

If f (x) = - 8A square x square + 6A third power X third power - 4A fourth power X fourth power, find f (x) divided by (- 4ax Square)

If f (x) = - 8A squared x squared + 6A cubic X Cubic - 4A quartic x quartic,
=-2ax^2(4a-3a^2x+2a^3x^2)
F (x) divided by (- 4ax squared)
=-2aX ^ 2 (4a-3a ^ 2x + 2A ^ 3x ^ 2) divided by (- 4ax squared)
=(4a-3a^2x+2a^3x^2)/2
=2a-3/2a^2x+a^3x^2

Simplification: 1 / 6a-1 / 2 (a + 1) + 1 / 3 (A-1) =

1／6a－1／2（a＋1）＋1／3（a－1）
=[(a+1)(a-1)-3a(a-1)+2a(a+1)]/6a(a+1)(a-1)
=(a²-1-3a²+3+2a²+2)/6a(a+1)(a-1)
=4/6a(a²-1)
=2/3(a³-a)

Simplify 5 + [6a-2 (A-1)] to

5 + [6a-2 (A-1)] is reduced to 7 + 4a

Simplify and evaluate a ^ 2-1 of a ^ 2 + 6A + 9 △ a + 1 * a ^ 2-9 of A-1, where a = 1

The original formula = (a + 1) (A-1) / (a + 3) & # 178; × 1 / (a + 1) × (a + 3) (A-3) / (A-1)
=(a-3)/(a+3)
When a = 1
Original formula = (1-3) / (1 + 3)
=-1/2

Simplify (a ^ 2-6a + 5) (a ^ 2 + 6A + 5) - A ^ 2 (a ^ 2-4)

There's a square difference ahead
（a^2-6a+5)(a^2+6a+5)-a^2(a^2-4)
=(a^2+5)^2-(6a)^2-a^4+4a^2
=(a^2+5)^2-20a^2-a^4
=a^4+10a^2+25-20a^2-a^4
=-10a^2+25

Decomposition factor: A & sup2; + B & sup2; + C & sup2; - 2BC + 2ac-2ab

It's the sum of squares formula
a²+b²+c²-2bc+2ac-2ab
=a²+(-b)²+c²+2ac+2a(-b)+2(-b)c
=(a-b+c)²