 # Given x-2y = 3, x2-2xy + 4y2 = 11, find the values of the following formulas: （1）xy； （2）x2y-2xy2．

## Given x-2y = 3, x2-2xy + 4y2 = 11, find the values of the following formulas: （1）xy； （2）x2y-2xy2．

（1）x2-2xy+4y2=（x-2y）2+2xy=11，
Substitute x-2y = 3 to obtain: xy = 1;
（2）∵xy=1，x-2y=3，
∴x2y-2xy2=xy（x-2y）=1 × 3=3．

### X square + y square + 2XY parts x square - 4Y square divided by x square + XY parts 2Y + X

X square + y square + 2XY parts x square - 4Y square divided by x square + XY parts 2Y + X
=(x-2y)(x+2y)/(x+y) ² · x(x+y)/(x+2y)
=x(x-2y)/(x+y)

### Evaluation: y ^ n (y ^ n + 9y-12) - 3 (3Y ^ n + 1-4y ^ n), where y = - 3, n = 2

y^n(y^n+9y-12)-3(3Y^n+1-4y^n)
=y^2n+9y^(n+1)-12y^n-9y^(n+1)+12y^n
=y^2n
=(-3)^(2 × 2)
=81

### Let x, y ∈ R, then the square of (3-4y-cosx) + (4 + 3Y + SiNx) is the least square,

∵ (3-4y-cosx) 2 + (4 + 3Y + SiNx) 2 = ([(34y) cosx] 2 + [(4 + 3Y) (SiNx)] 2) 2, analogizing the distance formula between two points | ab | = (x1x2) 2 + (y1y2) 2, and 3 (3-4y) + 4 (4 + 3Y) - 25 = 0, the formula obtained is a point on the straight line 3x + 4y-25 = 0 to a point on the circle x 2 + y 2 = 1

### Let x, y ∈ R, then the minimum value of (3-4y-cosx) 2 + (4 + 3Y + SiNx) 2 is () A. 4 B. 5 C. 16 D. 25

∵ (3-4y-cosx) 2 + (4 + 3Y + SiNx) 2 = ([(3-4y) - cosx] 2 + [(4 + 3Y) - (- SiNx)] 2) 2, analogizing the distance formula between two points | ab | = (x1-x2) 2 + (y1-y2) 2, and 3 (3-4y) + 4 (4 + 3Y) - 25 = 0, the formula is from a point on the straight line 3x + 4y-25 = 0 to a point on the circle x2 + y2 = 1

### Let x, y ∈ R, then the minimum value of (3-4y-cosx) 2 + (4 + 3Y + SiNx) 2 is () A. 4 B. 5 C. 16 D. 25

∵（3-4y-cosx）2+（4+3y+sinx）2=(
[(3-4y)-cosx]2+[(4+3y)-(-sinx)]2)2，
Analogy distance formula between two points | ab|=
(x1-x2)2+(y1-y2)2，
And 3 (3-4y) + 4 (4 + 3Y) - 25 = 0,
The formula obtained is the square of the distance from a point on the straight line 3x + 4y-25 = 0 to a point on the circle x2 + y2 = 1,
It can be seen from the drawing that the vertical section of 3x + 4y-25 = 0 is made through the origin o (0, 0), and the vertical foot is p, |op| = |3 × 0+4 × 0-25|
32+42=5，
The intersections of OP and circle are m and N respectively,
Obviously, the minimum value of (3-4y-cosx) 2 + (4 + 3Y + SiNx) 2 is | pm| 2 = (| op| - | om|) 2 = (| op| - 1) 2 = 16
Therefore, C