Simplification: (1) The square of 3 (x + 1) - 5 (x + 1) (x-1) + 2x (x-1) (2) (2x-5) (2x + 1) - (2x-3) (2x + 3)

Simplification: (1) The square of 3 (x + 1) - 5 (x + 1) (x-1) + 2x (x-1) (2) (2x-5) (2x + 1) - (2x-3) (2x + 3)

(1)
=3(x^2+2x+1)-5(x^2-1)+2x^2-2x
=3x^2+6x+3-5x^2+5+2x^2-2x
=4x+8
(2)
=4x^2+2x-10x-5-(4x^2+6x-6x-9)
=-8x+4

What's the answer to homework 3.1 in the first volume of junior high school mathematics?

a+5=8
1/3b=9
2x+10=18
1/3x-y=6
3a+5=4a
1/2b-7=a+b
a+b=b+a
ab=ba
a(b+c)=ab+ac
a+(b+c)=(a+b)+c
Book P82, 83
a=b
33,8,1,1

It is known that M is the root of the equation 2x ^ 2 + 3x-12 = 0, and the value of 1 / 3M ^ 2 + 1 / 2m is

2x^2+3x-12=0
2x^2+3x=12
1/3m^2+1/2m=(2x^2+3x)/6
=12/6
=2

Calculation problem, the best process, the answer is OK. X (x + 1) + (2-x) (2 + x), (a + 1 / 2b) (A-1 / 2b) - (3a-2b) (3a + 2b)

（1）1/3（6x-3）-1/2（8x-4）=5 （2）x/2-2x-4/3=1+5x+12/6 （3）3y-1/2-5y-7/6=1 （4）8x+7+2x=1+11x-6
（5）7（2x-1）-3（4x-1）=5（3x+2）-1 （6）3x-1/2=4x+2/5-1 （7）3/2（x+1）-x+1/6=1

1/3（6x-3）-1/2（8x-4）=5
2x-1-4x+2=5
-2x=4
x=-2
x/2-2x-4/3=1+5x+12/6
x/2-2x-4/3=1+5x+2
Multiply both sides by six at the same time, and you get
3x-12x-8=6+30x+12
-39x=26
x=-2/3
3y-1/2-5y-7/6=1
-2y=8/3
y=-4/3
8x+7+2x=1+11x-6
-x=-12
x=12
7（2x-1）-3（4x-1）=5（3x+2）-1
14x-7-12x+3=15x+10-1
-13x=13
x=-1
3x-1/2=4x+2/5-1
-x=-1/10
x=1/10
3/2（x+1）-x+1/6=1
Multiply both sides by six at the same time, and you get
9x+9-6x+1=6
3x=-4
x=-4/3

It is proved that if matrix A is not invertible, then the determinant of adjoint matrix is 0

First of all, if a = O, it is easy to see that a * = O, naturally there is | a * = 0. Suppose a ≠ o, a is irreversible, then | a | = 0, because AA * = | a | e, so AA * = O (0 matrix). Here we use a conclusion that the product of matrix is O: if AB = O, then R (a) + R (b) ≤ n. therefore, R (a) + R (a *) ≤ n, R (a) ≥ 1 is known from a ≠ o, so r (a *) ≤ n-1, that is, a * is not full rank, so | a * = 0

4x-7 = 5x + 2 to solve the equation

4x-7=5x+2
-7-2=5x-4x
x=-9

The solution of equation 2 + [X-1] [x + 3] = 0 has () absolute values

Discussion
1. When x

The number of real number pairs (x, y) satisfying the equation 11x ^ 2 + 2XY + 9y ^ 2 + 8x-12y + 6 = 0 is ()
The options are a.1, B.2, C.3 and D.4

Because 11x ^ 2 + 2XY + 9y ^ 2 + 8x-12y + 6 = 0 has a real root
So 11x ^ 2 + 2 (y + 4) x + (9y ^ 2-12y + 6) = 0 △≥ 0
That is, 4 (y + 4) ^ 2 - 44 * (9y ^ 2-12y + 6) ≥ 0
The solution is: (7y-5) ^ 2 ≤ 0,
So y = 5 / 7 (y has a unique value)
So the number of root pairs (x, y) satisfying the equation 11x ^ 2 + 2XY + 9y ^ 2 + 8x-12y + 6 = 0 is one
A should be chosen

We decompose the following formulas: (1) 3a square-6a; (2) the third power of 6A square B + the square of 10ab, and the third power of c-4ab
(3) 24x square - 8xy + 4x; (4) the square of a - 4Y square; (5) the fourth power of a - 2A square; the fourth power of b square + B
(6) 64M square n square - (m square + 16N Square) square

: (1) 3a squared-6a; = 3A (A-2) (2) 6A squared B to the third power + 10ab squared c-4ab to the third power = 2Ab & # 178; (3AB + 5bc-4b) (3) 24x squared-8xy + 4x; = 4x (6x-2y + 1) (4) a squared-4y squared; = (a + 2Y) (a-2y) (5) a to the fourth power - 2A squared B squared + B to the fourth power = (A & # 178; - B &