If the quadratic trinomial 2x × x + KX-5 (k is an integer) can be factorized in the range of integers, please write the possible value of K

If the quadratic trinomial 2x × x + KX-5 (k is an integer) can be factorized in the range of integers, please write the possible value of K

3,9,-3,-9

(x + 1) (x + 2) + 1 / 4 factorization

(x+1)(x+2)+1/4
=x²+3x+2+1/4
=x²+3x+9/4
=(x+3/2)²

-How to calculate 5x & # 178; + 10xy-25xy & # 178; factorization factor

The original formula = - 5x (x-2y + 5Y & # 178;)

Factorization 5x & # 178; + 10xy & # 178; - 15xy

Original formula = 5x (x + 2Y & # 178; - 3Y)

(- 1 / 4) x ^ 4 + 9y ^ 2 factorization factor writing process

(-1/4)x^4+9y^2
=9y^2-1/4x^4
=(3y)^2-(1/2x^2)^2
=(3y-1/2x^2)(3y+1/2x^2)

Group decomposition method: A-A ^ 3 + 4A ^ 2b-4ab ^ 2; 4x ^ 2-y ^ 2 + 4y-4

a-a^3+4a^2b-4ab^2
=a[1－﹙a²－4ab＋4b²﹚]
=a[1－﹙a－2b﹚²]
=a﹙1＋a－2b﹚﹙1－a＋2b﹚
4x^2-y^2+4y-4
=4x²－﹙y－2﹚²
=﹙2x＋y－2﹚﹙2x－y＋2﹚

Using the collocation method to solve x ^ 2 + x-3,4x ^ 2 + 8x-5, using the substitution method to solve (a-2b) ^ 2 + (a-2b) - 12, (x + y) ^ 2-4 (x + Y-1) factorization factor A ^ 2-B ^ 2 + 4A + 2B + 3

⑴x2+x-3= (x+2)(x-1) 4x2+8x-5= (2x+5)(2x-1)
⑵（a-2b)2+（a-2b)-12,(x+y) 2-4(x+y-1)
Let (a-2b) = x, then the original formula = x2 + X-12 = (x-3) (x + 4)
⑶a2-b2+4a+2b+3
=（a+b）(a-b)+4a+2b+3

Finding the integer solution of the equation xy = 2x + 2Y + 7

xy-2x-2y=7
x(y-2)-2(y-2)=3
(x-2)(y-2)=3
Integer solution
X-2 = - 1, Y-2 = - 3 or
X-2 = - 3, Y-2 = - 1 or the exchange of X and Y
We obtain x = 1, y = 1 or x = - 1, y = 1 or x = 1, y = - 1

2x-3=y;xy=209
The teacher said his birthday is on the y day of the month X. x is less than or equal to 12; y is less than or equal to 31. Tat

2x-3=y
xy=209
x(2x-3)=209
2x²-3x=209
2x²-3x-209 =0
x=0.25(3±√(9+8*209)
x=0.25(3±41)
x1=11
x2=-9.5
y1=22-3=19
y2=-19-3=-22
There are two sets of equations (11,19), (- 9.5, - 22)

Given y = 2 √ 2x-1 + 3 √ 1-2x + 4, find the value of XY

2x-1>=0
1-2x>=0
∴x=1/2
Substitute x = 1 / 2 into y = 2 √ 2x-1 + 3 √ 1-2x + 4 to get
y=4
∴xy=1/2×4=2