It is known that P: three numbers 2 ^ x, 2 / 2 ^ x, (1 / 2) ^ X are equal proportion sequence; Q: three numbers lgx, LG (x + 1), LG (x + 3) are equal difference sequence, then p is Q Then, what is the condition for P to be q? And the second question: we know the univariate quadratic equation of X: ① MX ^ 2-4x + 4 = 0: ② x ^ 2-4mx + 4m ^ 2-4m-5 = 0 (m ∈ z), and find the necessary and sufficient conditions for both equations to have integer solutions

It is known that P: three numbers 2 ^ x, 2 / 2 ^ x, (1 / 2) ^ X are equal proportion sequence; Q: three numbers lgx, LG (x + 1), LG (x + 3) are equal difference sequence, then p is Q Then, what is the condition for P to be q? And the second question: we know the univariate quadratic equation of X: ① MX ^ 2-4x + 4 = 0: ② x ^ 2-4mx + 4m ^ 2-4m-5 = 0 (m ∈ z), and find the necessary and sufficient conditions for both equations to have integer solutions

2 ^ x, 2 / 2 ^ x, (1 / 2) ^ X are in equal proportion sequence
Then 2 ^ x * (1 / 2) ^ x = (2 / 2 ^ x) ^ 2
(2/2^x)^2=1
∴ x=1
(1/2)÷（2/2）=1/2
q=1/2
The three numbers lgx, LG (x + 1) and LG (x + 3) form an arithmetic sequence
Then lgx + LG (x + 3) = 2lg (x + 1)
x(x+3)=(x+1)^2
3x=2x+1
x=1
1+1-1=1
p=1
Then p is a necessary and sufficient condition for Q
① MX ^ 2-4x + 4 = 0: integer solution
（x-2)^2+(m-1)x^2=0
(x-2)^2=(1-m)x^2>=0
|x-2|=±√（1-m) x
Then 1-m > = 0 and is an integer square (1)
②x^2-4mx+4m^2-4m-5=0（m∈Z）,
（x-2m)^2=4m+5>=0
m>-5/4 （2）
Synthesis (1), (2)
Then M = 1

Given that A-B = 2 root 3, ab = root 3, then the value of (a + 1) (B-1) is?

(a + 1) (B-1) = AB-A + B-1 = ab - (a-b) - 1 = root 3-2 root 3-1 = - root 3-1

Given that A-B = 2 times root 3-1, ab = root 3, find the value of (A-1) (B-1)

a-b=2√3-1>0
ab=√3>0
Then: a > b > 0
Then: (a + b) ² = (a-b) ² + 4AB
=(2√3-1)²+4√3
=13
a+b=√13
(a-1)(b-1)=ab-(a+b)+1
=√3-√13+1

Given: a + B = 3, ab = 1, find the value of (a + 1) (B + 1)

∵a+b=3
ab=1
A + B
=(a + b) - 2Ab
=9-2
=7
(AB) side
=1
(Party A + 1) (Party B + 1)
=Party A, Party B + Party A, Party B + 1
=(AB) + (a + b) + 1
=1+7+1
=9
The answer is 9
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Given a + B = 3, A-B = 2, find the value of a ^ 2 + B ^ 2, AB respectively

a^2+b^2=（a+b）^2-2ab=(a-b)^2+2ab
9-2ab=4+2ab
4ab=5
ab=5/4

Given a = 4 + root 3, B = 4-root 3, find the value of a / (a-root AB) - root B / (root a + root b)

A = 4 + √ 3, B = 4 - √ 3, a + B = 8, A-B = 2 √ 3A / (a-root AB) - root B / (root a + root b) = (√ a) ^ 2 / [(√ a) ^ 2 - √ a √ b] - B (√ a - √ b) / (√ a + √ b) = √ A / (√ a - √ b) - (√ ab-b) / (√ a-b) = √ a (√ a + √ b) / (√ a + √ b) (√ a - √ b) - (√ ab-b) / (a-b) = √ a + √ b) / (a - √ a - √ b) = (a + √ b) / (a-b) / (a - √ a - √ b) = (a - √ b) / (a -

It is known that four real numbers - 9, A1, A2 and - 1 are equal difference sequences, and five real numbers - 9, B1, B2, B3 and - 1 are equal proportion sequences, then B2 (a2-a1) = ()
A. 8B. -8C. ±8D. 98

From the question, we get a & nbsp; 2 − A & nbsp; 1 = D = − 1 + 94 − 1 = 83, B22 = 9, and because B2 is the third term in the equal ratio sequence, it has the same sign with the first term, that is, B2 = - 3  B2 (a2-a1) = - 8

If we know that 1, A1, A2 and 4 are equal difference sequences and 1, B1, B2, B3 and 4 are equal ratio sequences, then a1 + a2b2 equals ()
A. 12B. 2C. 52D. 3

∵ 1, A1, A2, 4 are equal difference sequence, ∵ A2 + A1 = 1 + 4 = 5, 1, B1, B2, B3, 4 are equal ratio sequence, ∵ B22 = b1b3 = 1 × 4 = 4, the solution is B2 = ± 2, B12 = B2 > 0, ∵ B2 = 2, ∵ a1 + a2b2 = 52

If 1, A1, A2, A3 and 4 are equal difference sequence and 1, B1, B2, B3 and 4 are equal ratio sequence, then a1 + A2 / B2
detailed

4=1+4d, 4=1*q^4
D = 3 / 4, q = root 2
a1=7/4, a2=10/4,b2=2
a1+a2/b2=7/4+10/4/2=3

-If 1, A1, A2, A3, - 4 become arithmetic sequence, - 1, B1, B2, B3, - 4 become arithmetic sequence, then the value of (a3-a1) / B2

A2 = (- 1-4) / 2 = - 2.5 A1 = (- 1 + A2) / 2 = - 1.75 A3 = (- 4 + A2) / 2 = - 3.25 B2 = (- 1) × (- 4) B2 = ± 2, so (a3-a1) / B2 = - 1.5 ± (± 2) = ± 0.75