(a square / A-1) - A-1, calculation problem

(a square / A-1) - A-1, calculation problem

The original formula = A & # 178; / (A-1) - (a + 1)
=[a²-(a+1)(a-1)]/(a-1)
=(a²-a²+1)/(a-1)
=1/(a-1)

Each of the following graphs is a triangle shaped pattern composed of several potted flowers. Each side (including two vertices) has n (n > 1) potted flowers. The total number of pots of each pattern is s. according to this rule, the relationship between S and N is___ ．

When n = 2, s = 1 + 2 = 12 × 2 (2 + 1) = 3; when n = 3, s = 1 + 2 + 3 = 12 × 3 (3 + 1) = 6; when n = 4, s = 1 + 2 + 3 + 4 = 12 × 4 × (4 + 1) = 10 When n = n, s = n (n + 1) 2

Knowledge of scientific plants in Grade Seven
In the structure of a flower, it is ()
A. Petal B. calyx C. receptacle D. stamen
After the opening of red plum blossom, the red part we can see is mainly ()
A. Stamen B. pistil C. petal D. sepal
Among the structures of the following plants, the fruit belongs to
1. Corn kernel 2. Watermelon seed 3. Peanut kernel 4. Rapeseed 5. Sunflower seed

1. D. stamens
2. C. petals
3. Peanut kernel
I'm from the biology book of the first semester of junior high school. Believe it or not! I don't know whether it's unit 2 or unit 3

First simplify, then evaluate: (3x2y-xy2) - 3 (x2y-2xy2), where x = 12, y = − 13

The original formula = 3x2y-xy2-3x2y + 6xy2 = 5xy2, when x = 12, y = − 13, the original formula = 5 × 12 × (− 13) 2 = 518

1. A new rectangular swimming pool, 50m long and 30m wide, has been built in new century primary school;
The plan drawn with scale () is the smallest. A.1:1000 b.1:1500 c.1:500

The plan drawn with scale (c.1:500) is the largest and the plan drawn with scale (b.1:1500) is the smallest

English story short, must be short, the best 2-6 sentences, to humor, because I want to recite down, now I want to

1.Little Robert asked his mother for tow cents.”What did you do with the money I gave you yesterday?”“I gave it to a poor old woman,”he answered.“You’er a good boy,”said the mother proudly.”He...

Why is the square difference of two consecutive odd numbers a mysterious number?

The mysterious number is the square difference of two continuous even numbers, that is, (2n + 2) ² - (2n) ² = 4N & #178; + 8N + 4-4n & #178; = 8N + 4 = 4 (2n + 1), that is to say, the mysterious number is four times the odd number. The square difference of two continuous odd numbers is (2n + 3) ² - (2n + 1) ² = 4N & #178; + 12n + 9-4n & #178; - 4

How many centimeters of wire is needed to make a rectangular frame with a bottom area of 36 square centimeters and a height of 4.5cm?

36 = 6 square 6 * 8 + 4.5 * 4 = 66 (CM)
A: at least 66 cm long wire is required

What time is it now

12:40:25

Two circles x2 + y2-4by-1 + 4B2 = 0 and X2 + Y2 + 2aX + A2-4 = 0. There are exactly three common tangents. If a and B belong to R and ab ≠ 0, what is the minimum value of A2 + B

Seeing that you asked the same question, I answered it shamelessly again,
Answer: minimum - (3 √ 2) / 2
Let's start with the formula of the two circle equation
X²+Y²-4by-1+4b²=0
x²+(y²-4by+4b²)=1
x²+(y-2b)²=1²
So this circle is a circle with (0,2b) as its center and 1 as its radius
X²+y²+2ax+a²-4=0
(x²+2ax+a²)+y²=4
(x+a)²+y²=2²
So this circle has (- A, 0) as its center and 2 as its radius
Because two circles have exactly three common tangent lines,
So it's easy to know that two circles are circumscribed
So the distance between the centers of two circles = the sum of the radii of two circles
That is √ [[0 - (- a)] & sup2; + (2b-0) & sup2;] = 1 + 2
√(a²+2b²)=3
a²+2b²=9
So the topic becomes familiar to us: "given a & sup2; + 2B & sup2; = 9, find the minimum of a & sup2; + B"
Substituting a & sup2; = 9-2b & sup2; into S = A & sup2; + B
We obtain s = 9-2b & sup2; + B
=-2(b²-½b)+9
=-2[b²-½b+1/16-1/16]+9
=-2[b-(1/4)]²+1/8+9
=-2[b-(1/4)]²+73/8
And because a is a real number
So a & sup2; = 9-2b & sup2; ≥ 0
So B & sup2; ≤ 18 / 4
-(3√2)/2≤b≤(3√2)/2
From the knowledge of quadratic function, when B = - (3 √ 2) / 2, s takes the minimum value
Here a = 0
So smin = 0 - (√ 18) / 2 = - (3 √ 2) / 2