Given a + B = - 2, A-B = 2, decompose (a 2 + B 2-1) 2-4 A 2B 2 first, and then evaluate

Given a + B = - 2, A-B = 2, decompose (a 2 + B 2-1) 2-4 A 2B 2 first, and then evaluate

The original formula = (A2 + b2-1-2ab) (A2 + b2-1 + 2Ab) = (a + B + 1) (a + B-1) (a-b + 1) (a-b-1) substituting a + B = - 2, A-B = 2, the original formula = 9

Square of factoring factor 2-8 (a-b)

2-8(a-b)²
=2[1-4(a-b)²]
=2[1+2(a-b)][1-2(a-b)]
=2(1+2a-2b)(1-2a+2b)

Fill in the appropriate length units in brackets. (1), 1 () - 9（
Smart house
Fill in the appropriate units of length in brackets
(1) , 1 () - 9 () 2 1 ()
(2) , 1 () - 9 () 2 1 ()
(3) , 1 () - 99 () 2 1 ()
(4) , 1 () - 999 () 2 1 ()

Meter, decimeter, decimeter

The two vertices of the ellipse e with focal length 2 are a and B respectively, and the vector AB and the vector n = (√ 2, - 1) are collinear

Vector AB and vector n = (√ 2, - 1) are collinear, indicating that a / b = radical 2
That is, a ^ 2 = 2B ^ 2
And 2C = 2, C = 1
c^2=a^2-b^2=b^2=1
a^2=2
So the elliptic equation is x ^ 2 / 2 + y ^ 2 = 1

Use 3 square with side length of 2 decimeters to form a rectangle. What are the perimeter and area of the rectangle?

The rectangle is 6 decimeters long and 2 decimeters wide
Perimeter = 2 * (2 + 6) = 16 decimeters
Area = 2 * 6 = 12 square decimeters

Given that the sum of solutions X and y of the equations 3x + 5Y = K + 2,3x + 3Y = k is equal to 2, then k = ()
Request detailed explanation, thank you!
Binary linear equation solution, thank you!

It is known that x + y = 2, x = 2-y
The known equation becomes
3(2-Y)+5Y=K+2
3(2-Y)+3Y=K
After decomposition
6-2Y=K+2
6=K
We get y = 1, and then substitute it into the equation x + y = 2 to get x = 1
The final solution is: k = 6, x = 1, y = 1

(1 / 2) it is known that the center of ellipse C is at the origin, the focus is on the x-axis, the left and right focus are F1F2, and the absolute value of F1F2 is 2, and the point (3 / 2) is at
(1 / 2) it is known that the center of ellipse C is at the origin, the focus is on the x-axis, the left and right focus are F1F2, and the absolute value of F1F2 is 2, and the point (3 / 1,2) is on ellipse C
(1) Finding the square of ellipse C

If F1F2 = 2, then C = 1F1 (- 1,0), F2 (1,0) point P (1,3 / 2) Pf1 = 5 / 2, PF2 = 3 / 2pf1 + PF2 = 2A = 4. So: if a = 2, then B & # 178; = A & # 178; - C & # 178; = 3. So, the equation of ellipse C is: X & # 178 / / 4 + Y & # 178 / / 3 = 12, let a (x1, Y1), B (X2, Y2), then s △ af2b = | y1-y2 | the other half is continued

Any prime number greater than 6 divided by 6 must have a remainder. What is the remainder?
Come on

Only 1 or 5
The remaining 2, 3 and 4 are not prime numbers

Given the function f (x) = 3 / X (1) prove that the function f (x) is a decreasing function on [1, + ∞ (2) find the function f (x) on [2, + ∞)+
The known function f (x) = 3 / X
(1) it is proved that the function f (x) is a decreasing function on [1, + ∞)
(2) find the function of F (x) on [2, + ∞)

(1)
Let a>b>=1 then a-b>0
f(a)-f(b)=3/a-3/b=3(b-a)/ab=2

Given that the odd function f (x) is a function on R, and f (1) = 2, f (x + 3), find f (8)
Given that the odd function f (x) is a function on R, and f (1) = 2, f (x + 3) = f (x), find f (8)

Since f (x) is an odd function over R, then
f(-x)=-f(x)
From F (1) = 2, we can see that f (- 1) = - 2;
From F (x + 3) = f (x), f (- 1) = f (2) = f (5) = f (8) = - 2 is obtained