Given a: 3 = 7: B, then the value of AB is () The sooner, the better. It's tonight. It's not more than 10:30!

Given a: 3 = 7: B, then the value of AB is () The sooner, the better. It's tonight. It's not more than 10:30!

a:3=7:b
ab=3*7
ab=21

Find the value of the formula A / | a | + B / | B | + AB / | ab |

1) When a > 0, b > 0
Original formula = 1 + 1 + 1 = 3
2）a>0,b

Given a + B = - 5, ab = 4, find the value of the following formula
Note that ^ 2 is the square
a^2 - b^2 =
Thank you,very much!

You didn't make the question clear just now
(a-b)^2=(a+b)^2-4ab=25-16=9
So (a-b) ^ 2 = 9
A-B = 3 or - 3
And a + B = - 5
When A-B = 3
2a=-2
a=-1
Now a ^ 2-B ^ 2 = 1-16 = - 15
When A-B = - 3
2a=-8
a=-4.
In this case, a ^ 2-B ^ 2 = 16-1 = 15

Given a = 2 + 3, B = 2 − 3, find the value of A2 + b2-3ab

∵a＝2+3，b＝2−3，∴a+b=4，ab＝(2+3)(2−3)＝4−3＝1，∴a2+b2-3ab=（a+b）2-5ab=42-5×1=11．

Given that a, B and C are three sides of triangle, and a ^ 2 + 4ac + 3C ^ 2-3ab-7bc + 2B ^ 2 = 0, prove 2B = a + C

a^2+4ac+3c^2-3ab-7bc+2b^2=0
(a+c)(a+3c)-3ab-7bc+2b²=0
(a+c)(a+3c)-(3a+7c)b+2b^2=0
(a+c-2b)(a+3c-b)=0
2b=a+c

It is known that a, B and C are the three sides of triangle, and a ^ 2 + 4ac + 3C ^ 2-3ab-7bc + 2B ^ 2 = 0

a^2+4ac+3c^2-3ab-7bc+2b^2=0
(a+c)(a+3c)-3ab-7bc+2b²=0
(a+c)(a+3c)-b(a+c)-2b(a+3c-b)
(a+c)(a+3c-b)-2b(a+3c-b)
(a+c-2b)(a+3c-b)=0
The three sides of a, B and C are triangles
∴a+c＞b
∴a+3c＞b
∴a+c=2b

When the square of ab-2 + (B + 1) = 0, find the value of 1 / AB + 1 / (A-1) (B-1) +. + 1 / (a-2011) (b-2011)!

When the square of ab-2 + (B + 1) = 0
b+1=0 b=-1
ab-2=0 a=-2
1/ab+1/(a-1)(b-1)+.+1/(a-2011)(b-2011)
=1/[(-1)x(-2)+1/[(-1-1)x(-2-1)]+…… +1/[(-1-2011)x(-2-211)]
=1/(1x2)+1/(2x3)+…… +1/(2012x2013)
=1-1/2+1/2-1/3+…… +1/2012-1/2013
=1-1/2013
=2012/2013

If | a + 2 | + b2-2b + 1 = 0, find the value of A2B + AB2

∫ a + 2 | + b2-2b + 1 = 0 | a + 2 | + (B-1) 2 = 0 | a = - 2, B = 1 | A2B + AB2 = AB (a + b) = (- 2) × 1 × (- 2 + 1) = 2, so A2B + AB2 = 2

If the square of (a + b) + B + 5 = B + 5, 2a-b-1 = 0, find the value of a and B

Because (a + b) & sup2; > = 0 and | B + 5 | > = 0
So B + 5 > = 0 means | B + 5 | = B + 5
So (a + b) & sup2; = 0
That is, a + B = 0
So a = - B
After substituting:
|2a-b-1|=|-2b-b-1|=|-3b-1|=0
-3b-1=0
b=-1/3
a=1/3

- 3 + (- 1) 0 square of 2011 SQUARE × (π - 3) + - 2 square of (1 / 2)

1 + (- 2) + root 27 - 5 - 2 times with 3 = with 3 - 6