If a and B satisfy a2b2 + A2 + B2 + 10ab + 16 = 0, the value of A2 + B2 is obtained The first two are squares,

If a and B satisfy a2b2 + A2 + B2 + 10ab + 16 = 0, the value of A2 + B2 is obtained The first two are squares,

a2b2+a2+b2+10ab+16
=（a2b2＋8ab＋16）＋（a2＋2ab＋b2）
＝（ab＋4）2＋（a＋b）2
＝0
So (AB + 4) 2 = 0 and (a + b) 2 = 0
So AB + 4 = 0 and a + B = 0
A = 2, B = - 2 or a = - 2, B = 2
So A2 + B2 = 4 + 4 = 8

Given a 2 + a 2b2-4ab + B 2 + 1 = 0, we can find the value of A-2003 × b-2003 + 1
2 and 2003 are power

First, I'll teach you to type the power of "^", shift + 6
a^2+a^2b^2-4ab+b^2+1=0
(a^2-2ab+b^2)+(a^2b^2-2ab+1)=0
(a-b)^2+(ab-1)^2=0
a=b,ab=1
(①a=1,b=1②a=-1,b=-1)
The above formula is reduced to (AB) ^ 2003 + 1
1^2003+1=2

Decomposition factor a2b2-4ab-a2-b2 + 1

a2b2-4ab-a2-b2+1
=(a2b2-2ab+1)-(a2+2ab+b2)
=(ab-1)2-(a+b)2
=(ab+a+b-1)(ab-a-b-1)

Excuse me: if any one of A1, B1, C1, D1 is equal to 5, and any one of A2, B2, C2, D2 is equal to 6, it will be displayed as "right"
Excuse me: if any number in A1, B1, C1, D1 is equal to 5, and any number in A2, B2, C2, D2 is equal to 6, it will be displayed as "right" in E2, otherwise it will be "wrong". What formula can be used to solve this problem?

=If (and (countif (A1: D1, "5") > 0, countif (A2: D2, "6"), "right", "wrong")

Prove inequality (A1 * B1 * C1 * D1) & #188; + (A2 * B2 * C2 * D2) & #188; ≤ ((a1 + A2) (B1 + B2) (C1 + C2) (D1 + D2)) & #188;

All variables in the question are positive numbers
First, we prove the following conclusions
Let X1 * x2 * X3 * X4 = m, then the minimum value of (1 + x1) (1 + x2) (1 + x3) (1 + x4) is reached when X1 = x2 = X3 = X4
It is proved that: 1. The minimum value exists, because when a certain Xi --- > infinity, (1 + x1) (1 + x2) (1 + x3) (1 + x4) > Xi --- > infinity, so its minimum value must reach at a certain point. Let this point be (a, B, C, d)
2. There must be a = b = C = D. if not, it is easy to verify that: (A1, B1, C, d) also satisfies a1b1cd = m, and (1 + A1) (1 + B1) (1 + C) (1 + D) = (1 + a) (1 + b) (1 + C) (1 + D) = (1 + (x1 * x2 * X3 * x4) ^ (1 / 4)) ^ 4
Let x 1 = a 1 / a 2, x 2 = B 1 / B 2, x 3 = C 1 / C 2, x 4 = D 1 / D 2

For example: a B C D 1 A1 B1 C1 D1 2 A2 B1 C1 D1 3 A3 B2 C2 C2 4 A4 B2 C2 C2
How can I delete 1 / 3 or 2 / 4?

Using VBA method to solve the problem, the code is as follows:
Sub remove duplicate ()
Sheets("sheet1").Select
i = 0
j = 0
k = 0
l = 0
m = 2
For i = 1 To 4
For j = 1 To 4
For k = m To 4
For l = 1 To 4
If Cells(k, l) = Cells(i, j) Then
Cells(k, l).Select
Selection.ClearContents
End If
Next l
Next k
Next j
m = m + 1
Next i
End Sub

Motor vehicle model: C1, C2, C3, C4

C1 license can drive: small car, small automatic car, low speed truck, tricycle
C2 driving license can only drive small automatic cars, other models can not drive
C3 driving license can only drive low speed trucks and tricycles, other models can not drive
C4 can only drive three wheelers

Window size, window specification C1, C2, C3, C4, C5, what's the specific size

C1, C2, C3, C4, C5. It's just the number. The specific window size and window specification should see the drawing description

How many terms are there after the expansion of product (a1 + A2 + a3) (B1 + B2 + B3 + B4) (C1 + C2 + C3 + C4)?
How do I get to 48
The answer is 60
What's going on?

(a1 + A2 + a3) (B1 + B2 + B3 + B4) has: 3 * 4 = 12 items
(a1 + A2 + a3) (B1 + B2 + B3 + B4) (C1 + C2 + C3 + C4) has 12 * 4 = 48 items

How many terms are there after the expansion of product (a1 + 12 + a3) (B1 + B2 + B3 + B4) (C1 + C2 + C3 + C4)

=(a1b1+a1b2+a1b3+a1b4+a2b1+a2b2+a2b3+a2b4+.+)(c1+c2+c3+c4)
=(a1b1c1+.+a3b4c4)
There are 3 * 4 * 4 = 48 items