Factorization of Y ^ 2 + 1 / 6y-1 / 6

Factorization of Y ^ 2 + 1 / 6y-1 / 6

The original formula = y & # 178; + 1 / 6y + 1 / 144-1 / 144-1 / 6
=(y+1/12)²-25/144
=(y+1/12+5/12)(y+1/12-5/12)
=(y+1/2)(y-1/3)

Given the circle C: x ^ 2 + y ^ 2-2x + 4y-4 = 0, ask if there is a line L1 with slope of 1, so that the circle m with the diameter of chord AB cut by circle C by line L1 just passes through the origin

Hypothesis exists
Straight line y = x + B
Substituting
2x²+(2b+2)x+b²+4b-4=0
x1+x2=-(b+1)=-b-1
x1x2=(b²+4b-4)/2
y=x+b
y1y2=x1x2+b(x1+x2)+b²=(b²+2b-4)/2
AB is the diameter, O is on the circle
So OA vertical ob
OA slope Y1 / x1, ob is Y2 / x2
So (Y1 / x1) (Y2 / x2) = - 1
y1y2=-x1x2
(b²+2b-4)/2=-(b²+4b-4)/2
b²+3b-4=0
b=-4,b=1
So X-Y-4 = 0 and X-Y + 1 = 0

Given the circle C: x2 + Y2 + 2x-4y-4 = 0, (1) if the line L passes through point a (1,0) and the chord length cut by circle C is 2, the equation of the line is obtained; (2) if the circle m passes through the center of circle C and is tangent to the line L in (1), if the center of circle m is on the line y = x + 1, the equation of circle m is obtained

(1) Let L: y = K (x-1) radius be 3, chord length be 2, and the distance between center C and l be 22; |Let m (a, a + 1), ∵ r = | a − (a + 1) − 1 | 2 = 2, M: (x-a) 2 + (y-a-1) 2 = 2, and C (- 1, 2) ∵ (- 1-A) 2 + (1-A) 2 = 2, then a = 0, so the equation of circle m is: x2 + (Y-1) 2 = 2

[x-2y] [x + 2Y] - the square of [x + 2Y] is equal to

[x-2y] [x + 2Y] - the square of [x + 2Y]
=（X-2Y）（X+2Y）-(X+2Y)²
=X² -4Y² -（X²+4XY +4Y²）
= -4XY -8Y²

If x + 2Y = 5, xy = 1, then 2x2y + 4xy2=______ ．

∵ x + 2Y = 5, xy = 1, ∵ 2x2y + 4xy2 = 2XY (x + 2Y) = 2 × 1 × 5 = 10, so the answer is: 10

Given x + y = 1, the value of 2x ^ 2 = 4xy + 2Y ^ 2 is

2x^2+4xy+2y^2
=2(x^2+2xy+y^2)
=2(x+y)^2
=2*1^2
=So the final answer is 2

Given 1 / X - / y = 2, find the value of X + 4xy-y / 2x-xy-2y
How do I figure it out = - 2 / 5

Because 1 / X-1 / y = 2x-y = 2XY
Then (x + 4xy-y) / (2x-xy-2y) = (X-Y + 4xy) / (2x-2y-xy) = (2XY + 4xy) / (2 * 2xy-xy) = 6xy / 3xy = 2

Given 1 △ Y-1 △ x = 5, find the value of (2x + 4xy-2y) / (x-3xy-y)

1/y - 1/x =5
So: X / XY - Y / xy = 5
That is: (X-Y) / xy = 5
x－y ＝5xy
（2x+4xy－2y）/ （x-3xy-y）
＝（（2x－2y）＋4xy ）/（（x－y）－3xy）
＝(2*5xy+4 xy)/(5xy－3xy）
=14 / 2
=7

If - x ^ 2Y = 2, what is the value of - XY {x ^ 5Y ^ 2-x ^ 3Y + 2x}

-x^2y=2
x^2y=-2
-xy{x^5y^2-x^3y+2x}
=-x^6y^3+x^4y^2-2x^2y
=-(x^2y)^3+(x^2y)^2-2x^2y
=8+4+4
=16

X ^ 5y-x ^ 3Y + 2x ^ 2Y XY factorization factor

x^5y-x^3y+2x^2y-xy
=xy(x^4-x^2+2x-1)
=xy[x^4-(x^2-2x+1)]
=xy[x^4-(x-1)^2]
=xy(x^2-x+1)(x^2+x-1)