Rectangle perimeter 16, length x width y, satisfy (X-Y) square - 2x + 2Y + 1 = 0, calculate area? If (X-Y) & sup2; - 2x + 2Y + 1 = 0, calculate the area

Rectangle perimeter 16, length x width y, satisfy (X-Y) square - 2x + 2Y + 1 = 0, calculate area? If (X-Y) & sup2; - 2x + 2Y + 1 = 0, calculate the area

Circumference: 2 (x + y) = 16 2x + 2Y = 16 reduction: x + y = 8 (X-Y) & sup2; - 2x + 2Y + 1 = (X-Y) & sup2; - 4xy-2x + 2Y + 1 because:x+y=8 ,2x+2y=16 so:8²-4xy-16+1=0 -4xy=-49 xy=12.25

If the length and width of a rectangle increase by 18 cm, the area will increase by 20 square cm. What is the perimeter of the original rectangle

Rectangle area = length multiplied by width
Then (L + 18) * (W + 18)
=L x W + 18 (L + W) + 18x18
Then 18x18 + 18x (length + width) = 20
There is no solution

The perimeter of a rectangle is 18cm. If its length and width increase by 3cm, how much will its area increase

Use the special case method to solve the problem
Suppose this rectangle is a special rectangle square, then
18 / 4 = 4.5 cm Side length
So (4.5 + 3) (4.5 + 3) - 4.5 × 4.5 = 36 square centimeters
That is, the area is increased by 36 square centimeters

3 & sup2; + 4 & sup2; = 5 & sup2;, 8 & sup2; + 6 & sup2; = 10 & sup2;, 15 & sup2; + 8 & sup2; = 17 & sup2

When n = 1, (1 * 3) ^ 2 + (2 * 2) ^ 2 = 5 ^ 2n = 2, (2 * 4) ^ 2 + (2 * 3) ^ 2 = 10 ^ 2n = 3, (3 * 5) ^ 2 + (2 * 4) ^ 2 = 17 ^ 2n = 4, (4 * 6) ^ 2 + (2 * 5) ^ 2 = 26 ^ 2n = 5, (5 * 7) ^ 2 + (2 * 6) ^ 2 = 37 ^ 2... N = x, (x * (x + 2)) ^ 2 + (2 * (x + 1)) ^ 2 = i.e. (x ^ 2 + 2x)

Put the nine numbers 123456789 in the following nine () to make the equation true
()\()=()()\()=()()\()()=2

(6)\(3)=(1)(8)\(9)=(5)(4)\(2)(7)=2

In the space rectangular coordinate system, if the plane passes through the point P (- 1,3,2), and the normal vector of the plane a = (2,1, - 2), then the points belonging to the plane among the following points are
A（2,-3,2） B（2,0,1）C(2,3,0) D(0,-3,2)

Vector PA = (3, - 6,0) vector PA dot times vector a = (3, - 6,0) ● 2,1, - 2) = 0, so point a is in the plane; vector Pb = (3, - 3, - 1) vector Pb dot times vector a = (3, - 3, - 1) ● 2,1, - 2) = 5, so point B is not in the plane; vector PC = (3,0, - 2) vector PC dot times vector a = (3,0

It is known that the asymptote equation of hyperbola is y = plus or minus 3 / 4x, and the eccentricity of hyperbola is calculated,
I calculated 4 / 5 and 3 / 5, but the answer is just the reverse of my denominator

Carelessness should be avoided as far as possible

If the product of two numbers is negative, and the sum is also negative, please write the two numbers that meet the requirements?

Two numbers that meet the requirements: - 2 and 1

Let log2 / 1 ^ (x + 2) power > log2 / 1 ^ x ^ 2 power, then the value range of X is? Final answer (- 2, - 1) U (2, positive infinity)

log(1/2)*(x+2)＞log(1/2)*x^2.
Log (1 / 2) * x is a decreasing function, so
X + 2 ＜ x ^ 2, the solution is x ＞ 2 or X ＜ - 1
At the same time, the true number is greater than 0, i.e
X + 2 ＞ 0, 2 ^ x ＞ 0. That is x ＞ - 2
Finding the union of X ＞ 2 or X ＜ - 1 and X ＞ - 2
(- 2, - 1) U (2, positive infinity)
Can I?

If the quadratic equation of one variable X-2 (1-m) x + m square = 0 has real roots a and B, then the value range of a + B is

∵ the equation x ^ 2-2 (1-m) x + m ^ 2 = 0 has two real roots a and B
∴△≥0
∴4（1-m）^2-4×m^2≥0
4-8m+4m^2-4m^2≥0
m≤1/2
∵a+b=2-2m
And m ≤ 1 / 2
∴2m≤1
∴-2m≥-1
∴2-2m≥1
That is, a + B ≥ 1