Xiaomin has 80 cube blocks with the edge length of 1cm. He wants to cut these blocks into a cube with the largest volume. He can use () blocks at most

Xiaomin has 80 cube blocks with the edge length of 1cm. He wants to cut these blocks into a cube with the largest volume. He can use () blocks at most

There are 80 cubes in total, so the volume cannot be greater than 80cm ^ 2
If the side length of a large cube is x, then there will be
X^3

English Assessment Handbook for the first semester of junior high school p35
The grammar of this unit is - and - (unit 2)
Their respective usages are

The syntax of this unit is subject from and object from
Their respective usages are 1. To be the subject and 2. To be the object
(the above is what our teacher said)

Factorization (1) (X & # 178; + 4) & # 178; - 16x & # 178; (2) (X & # 178; + 2x) & # 178; + 2 (X & # 178; + 2x) + 1

（1）（x²+4）²-16x²
=[(x²+4)+4x][(x²+4)-4x]
=(x+2)²(x-2)²
（2）（x²+2x）²+2（x²+2x）+1
=[(x²+2x)+1]²
=[(x+1)²]²
=(x+1)^4

Factorization (AX + 1) (AX + 2) (AX + 4) (AX + 5)

(AX + 1) (AX + 2) (AX + 4) (AX + 5) = (AX + 1) (AX + 5) (AX + 2) (AX + 4) = (a ^ 2x ^ 2 + 6AX + 5) (a ^ 2x ^ 2 + 6AX + 8), let a ^ 2x ^ 2 + 6AX + 5 = t, then: = t (T + 3) do it here, then you will find that you have a problem with this problem. The problem is originally the product of several terms, which has been factorized. Why do you need to decompose it again

X2 + ax-bx-by + XY AB factorization!

x²+ax-bx-by+xy-ab
=x(x+y)+a(x-b)-b(x+y)
=(x-b)(x+y)+a(x-b)
=(x-b)(x+y+a)

Factorization of X4 + AX3 - (2A2 + 1) x2 ax + 2A2
After the letter is the number
There is also the problem of 2x2 + 5xy-3y2-3x + 5y-2

(1)x4+ax3-(2a2+1)x2-ax+2a2
=x^4+ax^3-2a^2*x^2-x^2-ax+2a^2
=(x^4-x^2)+(ax^3-ax)-(2a^2*x^2-2a^2)
=x²(x²-1)+ax(x²-1)-2a²(x²-1)
=(x²-1)(x²+ax-2a²)
=(x+1)(x-1)(x+2a)(x-a)
(2)2x²+5xy-3y²-3x+5y-2
∵2x²+5xy-3y²=(2x-y)(x+3y)
Let 2x & # 178; + 5xy-3y & # 178; - 3x + 5y-2 = (2x-y + a) (x + 3Y + b) = 2x & # 178; + 5xy-3y & # 178; + (a + 2b) x + (B-3A) y + ab
∴a+2b=-3①,3a-b=5② ab=-2③
A = 1 and B = - 2 are obtained from (1) and (2) and substituted by (3),
∴2x²+5xy-3y²-3x+5y-2=(2x-y+1)(x+3y-2)

In the following deformation from left to right, the factorization is a. (x + 2) (X-2) = the square of x-4, B. the square of x-4 + 4x = (x + 2) (X-2) + 4x
C. Square of X-1 / y = (x + 1 / y) (x-1 / y)
D. Square of X - 1 / 2x + 1 / 16 = (x-1 / 4) square

A

Factorization ax ay + x ^ 2-y ^ 2

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

In the following formulas, the correct deformations are: 1, from x = y to AX = ay; 2, from AX = ay to x = y; 3, from x = y to x + a = y + A; 4, from X / a = Y / A to x = y

1 is wrong, the number of the same multiplication on both sides of the equation is not 0, multiplying by 0 is meaningless
2 is wrong, a is not equal to 0
Three is right
4 is wrong, a is not equal to 0

Given the equation AX = ay, the following deformation is correct ()
A. x=yB. 3-ax=3-ayC. ay=-axD. ax+1=ay-1

A. When a = 0, X is not necessarily equal to y, so a is wrong; B 3-ax = 3-ay, so B is correct; cay = ax, ay ≠ - ax, so C is wrong; DAX + 1 = ay + 1, so D is wrong; so B