Given that real numbers x and y satisfy (X-2) ^ 2 + (Y-1) ^ 2 = 1, find the maximum and minimum of Z = ((y + 1) / X

Given that real numbers x and y satisfy (X-2) ^ 2 + (Y-1) ^ 2 = 1, find the maximum and minimum of Z = ((y + 1) / X

Let x = 2 + Sina, y = 1 + cosa, let (y + 1) / x = (COSA + 1) / (Sina + 2) = aasina + 2A = cosa + 1asina cosa = 1-2a √ (a ^ 2 + 1) sin (a + b) = 1-2asin (a + B) = (1-2a) / √ (a ^ 2 + 1) from - 1 ≤ sin (a + b) ≤ 1, get - 1 ≤ (1-2a) / √ (a ^ 2 + 1) ≤ 1 (1-2a) ^ 2 / (a ^ 2 + 1) ≤ 1a (3a-4) ≤ 00 ≤

A number consists of seven ninths. This number is () and its reciprocal is ()

A number consists of seven ninths. This number is (7 / 9) and its reciprocal is (9 / 7)

Simplified evaluation: (2x & sup2; - y) (2x & sup2; + y) - (2Y + X & sup2;) (2y-x & sup2;) where x = - 1, y = - 2
Simplified evaluation: (2x & sup2; - y) (2x & sup2; + y) - (2Y + X & sup2;) (2y-x & sup2;) where x = - 1, y = - 2

(2x2-y)(2x2+y)-（2y+x2）（2y-x2）=4x^4-y^2-4y^2+x^4=5x^4-5y^2=5-5*4=-15

The function f (x) = - x2 + ax-b is known. If a and B are arbitrary numbers from the interval [0,4], then the probability of F (1) > 0 is ()
A. 916B. 932C. 716D. 2332

F (1) = - 1 + A-B > 0, i.e. A-B > 1, as shown in the figure, a (1,0), B (4,0), C (4,3), s △ ABC = 92, P = 924 × 4 = 932

① The solution of {y = two thirds x, 3x-4y = 2} the solution of {y = 1-x, 3x-17 = - y

① Substituting y = 2 / 3x
3x－8/3x＝2
1/3x=2
X = 6
y=4
② Substituting y = 1-x into
3x-17=-(1-x)
3x-17=x-1
3x-x-17+1=0
2x-16=0
2x=16
X = 8
y=-7

The image of quadratic function takes the straight line x = - 2 as the symmetry axis, the function has the minimum value - 4, and then passes through the point (0,1), the analytic expression of the function is obtained

Let the function be y = a (x + 2) ^ 2-4, and substitute (0,1) into 1 = a (0 + 2) ^ 2-4, a = 5 / 4,
The analytic formula of the function is: y = 5 / 4 (x + 2) ^ 2-4, y = 5x ^ 2 / 4 + 5x + 1

If the real number a B satisfies B & # 178; + a-2b + 2 = 0, then the value range of a is
If the real number a B satisfies B & # 178; + a-2b + 2 = 0, then the value range of a is
A. Less than or equal to - 1 B. greater than or equal to C. less than or equal to 1 D. greater than or equal to 1

a=-b^2+2b-2
=-(b^2-2b+1)+1-2=-(b-1)^2-1
When B = 1, take a and take the maximum value - 1
So a

If the equation (3-m) x ^ (| m | - 2) + 7 = 2 about X is a linear equation of one variable, is the value of m certain? Why?

The equation (3-m) x ^ (| m | - 2) + 7 = 2 is a linear equation with one variable
Then 3-m ≠ 0, | m | - 2 = 1
m≠3,|m|=3
m≠3,m=±3
So m = - 3

3.8x + 0.5 × 3 = 11 16 (x + 6) = 432
2 questions

3.8x=11-1.5
3.8x=9.5
x=2.5
x+6=432/16
x+6=27
x=21

The positions of the real numbers a and B on the number axis are shown in the figure, simplifying | a + B | + | A-B | + | b-a|
___ .________ .______ .____
a 0 b

___ .________ .______ .____
a 0 b
a