Given a + B = 3, ab = - 12, find the following values (1) (A-1) (B-1); (2) a ^ 2 + B ^ 2; (3) (a-b) ^ 2

Given a + B = 3, ab = - 12, find the following values (1) (A-1) (B-1); (2) a ^ 2 + B ^ 2; (3) (a-b) ^ 2

1、（a-1)(b-1)
=ab-a-b+1
=ab-(a+b)+1
=-12-3+1
=-14
2、a^2+b^2
=(a+b)^2-2ab
=9+24
=33
3、(a-b)^2
=(a+b)^2-4ab
=9+48
fifty-seven

Given: a + B = 3, ab = - 12, find the value of the following formula:
The second power of a - the second power of AB + B
The second power of a + the second power of B
The second power of (a-b)

The second power of a - AB + B = (a + b) &# 178; - 3AB = 3 & # 178; - 3 × (- 12) = 45
The second power of a + the second power of B = (a + b) &# 178; - 2Ab = 3 & # 178; - 2 × (- 12) = 33
The second power of (a-b) = (a + b) &# 178; - 4AB = 3 & # 178; - 4 × (- 12) = 57

Given a + B = 3, ab = - 12, find the value of the following formula
ab^2+a^2b

ab^2+a^2b
=ab(a+b)
=-12×3
=-36;
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Once when he was doing his homework, he took the divisor 40 as 4, and the result was that the quotient was 45 and the remainder was 3. So what's the correct quotient and remainder?

（45×4+3）÷40
=183÷40
=4······23

When he did a division problem, he took the divisor 30 as three, and the quotient of the result was 93. What was the correct quotient and remainder?

93X3＝279
279÷30＝9.9

If the divisor and divisor are reduced by 10 times at the same time, the quotient and remainder are ()
A. Shang 5 Yu 3B. Shang 50 Yu 3C. Shang 5 Yu 30d. Shang 50 Yu 30

If the divisor and divisor are reduced by 10 times at the same time, the quotient is still 50. Because the divisor is reduced by 10 times, the remainder is also reduced by 10 times to 3

In a division with remainder, the sum of divisor, divisor, quotient and remainder is 93, the quotient is 6, and the remainder is 3?

Divisor = (96-6-3-3) / (1 + 6) = 12
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(√ 12 - √ 20) * (√ 15 + 5) - ((√ 10 - √ 2) power 2)

(√12 -√20)×(√15+5)-(√10-√ 2)²
=√180+5√12-√300-5√20-10+2√20-2
=6√5+10√3-10√3-10√5+4√5-8
=-8

If every item of the equal ratio sequence {an} is positive and a10a11 + a9a12 = 2e5, then lna1 + lna2 + +lna20=______ ．

∵ sequence {an} is an equal ratio sequence, and a10a11 + a9a12 = 2e5, ∵ a10a11 + a9a12 = 2a10a11 = 2e5, ∵ a10a11 = E5, ∵ lna1 + lna2 + lna20=ln（a1a2… A20) = ln (a10a11) 10 = ln (E5) 10 = lne50 = 50

If the N-1 power of 15A is the same as the m-th power of the 3-th power of negative 5 / 4A, then what is M-N equal to

The N-1 power of 15A is similar to the m-th power of the 3-th power B of negative 5 / 4A
Then n-1 = 3
N=4
There is no B, so m = 0
m-n=0-4=-4