(2x 3Y) & # 178; + 2 (2x 3Y) - 4 = 0, then the value of 2x + 3Y is

(2x 3Y) & # 178; + 2 (2x 3Y) - 4 = 0, then the value of 2x + 3Y is

2X + 3Y as a whole, set as a,
The original equation is changed into:
A^2+2A=4
(A+1)^2=5,
A+1=±√5,
A = - 1 + √ 5, or a = - 1 - √ 5,
That is 2x + 3Y = - 1 + √ 5 or - 1 - √ 5

If | X-1 / 2x-3 | + (3Y + 1 / y + 4) & # 178; = 0, find the value of (3 / 2x-3) - (3 / 3y-1)

|x-1/2x-3|+（3y+1/y+4）^2=0
If the sum of two nonnegative numbers is 0, then both numbers are 0
(x-1)/(2x-3)=0
x=1
(3y+1)/(y+4)=0
y=-1/3
3/（2-3）-3/(-1-1)
=-3+3/2
=-3/2

（x+3）（x-3）（x²-9）；（2x+y）²-（2x+3y）（2x-3y）

Solution
(x+3)(x-3)(x²-9)
=(x²-9)(x²-9)
=(x²-9)²
=x^4-18x²+81
(2x+y)²-(2x+3y)(2x-3y)
=4x²+4xy+y²-(4x²-9y²)
=(4x²-4x²)+4xy+(y²+9y²)
=4xy+10y²

Factorization a ^ 4x ^ n + 2-4x ^ n (n is an integer)

a^4x^（n+2）-4x^n＝x^n[(a^4)x^2-4]=x^n[(a^2)x-2][(a^2)x+2]

It is known that the vector systems α 1, α 2 and α 3 are a basic solution system of the homogeneous linear equation system AX = 0
Then the fundamental solutions of the following vector systems that are not homogeneous linear equations AX = 0 are a, α 1 + α 2, α 2 + α 3,3 α 3 - α 1b, α 1 + α 2,2 α 3 - α 1, α 2 + α 3C, α 1 - α 3, α 2 + α 3, α 1 + 2 α 2 + α 3 D, α 1 - α 3, α 2 + 2 α 3,3 α 1 + α 2

If you can't see it through direct observation, you can calculate the determinant. It's not the basic solution system that equals 0
Such as (a) determinant=
1 1 0
0 1 1
-1 0 3
= 2
(B)
1 1 0
-1 0 2
0 1 1
=-1
(C)
1 0 -1
0 1 1
1 2 1
=0
Select (c)
In fact, (α 1 - α 3) + 2 (α 2 + α 3) - (α 1 + 2 α 2 + α 3) = 0

Party A and Party B cooperate in a batch of parts, which can be completed in 20 days. The work efficiency ratio of Party A and Party B is 5:4. How many parts does Party A and Party B complete each day

We are doing a long middle school problem, this is a primary school problem
The routine primary school practice is as follows:
If Party A and Party B cooperate for 20 days, then the sum of work efficiency of Party A and Party B is 1 / 20
(work efficiency ratio) a: B = 5:4, then a: (a + b) = 5 / 9
The efficiency of a is (1 / 20) * (5 / 9) = 1 / 36
The work efficiency of B is (1 / 20) - (1 / 36) = 1 / 45

Solution equation: x-25% = 0.5 3 / 4x / 3 = 1 / 2 7 / 8x-1 / 2x = 3 / 4

First equation:
x=0.5+25%
x=0.75
Second equation:
1/4x=1/2
x=2
The third equation:
3/8x=3/4
x=2

The two warehouses of a and B originally store 250 tons of goods. Now the warehouse of a carries another 36 tons, and the warehouse of B transports 1 / 3 of the goods. At this time, the weight of the goods stored in the two warehouses of a and B is equal. What is the original weight of the warehouse of a

If warehouse A has x tons, warehouse B will be: (x-36) / (1-1 / 3)
According to the meaning of the title:
X+（X-36）/（1-1/3）=250
X+（X-36）*3/2=250
X=121.6
That is, the original 121.6 tons of goods in warehouse a

f ' (-2)= - 4e^( - 3)+4e^( - 3)-12a-4b=0
f ' (1)=2+1-3a+2b=0
The solution is: a = 1 / 3 B = - 1 Why do the above two equations get a = 1 / 3 B = - 1

Two fifths of one meter is as long as one fifth of two meters

The same length, 1 times 5 / 2 = 0.4, 2 times 5 / 1 = 0.4