There are several different ways to make a large cuboid with 8 small cube blocks with the edge length of 2cm? For example, length: 16; width: 2; height: 2; volume: 64; surface area: 136

There are several different ways to make a large cuboid with 8 small cube blocks with the edge length of 2cm? For example, length: 16; width: 2; height: 2; volume: 64; surface area: 136

There are two kinds of spelling, one is to make a long strip of eight consecutive pieces, the other is to make four pieces on the left and four pieces on the right
The first spelling is as you said: length 16, width 2, height 2, volume 64, surface area 136
The second spelling is length 8 width 4 height 2 Volume 64 surface area 112
In fact, there is another spelling, but the result is cube, so it doesn't match the problem, so it doesn't count
The spelling is 4 below, 4 above, 8 in total

There are several different ways to make a large cuboid with 8 small cube blocks with the edge length of 2cm. What do you find?

Two species: length 2 × 4 = 8 cm, width 2 cm, height 2 cm, Volume 16 × 2 × 2 = 64 cubic cm, two species: length 2 × 4 = 8 cm, width 2 cm, height 2 × 2 = 4 cm, Volume 8 × 2 × 4 = 64 cubic cm, three species: length 2 × 2 = 4 cm, width 2 × 2 = 4 cm, height 2 × 2 = 4 cm, Volume 4 × 4 × 4 = 64 cubic cm

How to put 27 connected cube blocks together into a 3 × 3 cube
Come on
It's a cube like Rubik's cube

Three in a row... And then to the side of the fold... A face out... So the three face fold out

Factorization of (AX + by) ^ 2 + (ay BX) ^ 2 + C ^ 2x ^ 2 + C ^ 2Y ^ 2

（ax+by)^2+(ay-bx)^2+c^2x^2+c^2y^2
=a²x²+b²y²+a²y²+b²x²+c²x²+c²y²
=(a²+b²+c²)x²+(a²+b²+c²)y²
=(a²+b²+c²)(x²+y²)

Factorization of (AX + by) ^ 2 + (BX + ay) ^ 2 + C ^ 2x ^ 2 + C ^ 2Y ^ 2

=a2x2+2abxy+b2y2+a2y2+2abxy+b2x2+c2x2+c2y2 =(a2+b2+c2)(x2+y2)+4abxy

Decomposition factor: 2Y ^ 2-5xy + 2x ^ 2-ay-ax-a ^ 2

=(2x-y)(x-2y)-ax-ay-a²
=(2x-y+a)(x-2y-a)

If the polynomial x ^ 2 + 2XY + ay ^ 2-4 = (x + y + 2) (x + Y-2), then a =? If the polynomial x ^ 2-y ^ 2 + 4x + M can be factorized, write an M that conforms to the formula

If polynomial x ^ 2 + 2XY + ay ^ 2-4 = (x + y + 2) (x + Y-2)
(x+y+2)(x+y-2)=(x+y)^2-4=x^2+2xy+y^2-4 a=1
X ^ 2-y ^ 2 + 4x + M can be factorized,
x^2-y^2+4x+M
=(x^2-y^2)+4x+M
=(x+y)(x-y)+4x+M M=4y
=(x+y)(x-y)+4(x+y)
=(x+y)(x-y+4)
So m = 4Y

X ^ - 4x + 4 = 0; why (X-2) ^ = 0;
Which is the middle term, which is the prime minister and the last term?

The square formula of the difference between two numbers: (a-b) ^ 2 = a ^ 2-2ab-b ^ 2
Where a = x, B = 2
(x-2)^2 = x^2 - 2*x*2 + 2^2 = x^2-4x+4

How to solve (X-2) (2x + 1) = 1 + 2x by factorization

(x-2)(2x+1)=1+2x
(2x+1)(x-2-1)=0
(2x+1)(x-3)=0
X = - 1 / 2 or x = 3

How to solve the inequality x ^ - X-6 > 0 by factorization

X^-X-6 ＝ (x-3)*(x+2)
X^-X-6>0
(x-3)*(x+2)>0
So x-3 > 0 and X + 2 > 0 = > x > 3
Or x-3