Known: 3x2-xy-4y2 = 0 (x ≠ y) find (x-2y) / (x + 2Y) Known: the square of 3x - the square of xy-4y = 0 (x ≠ y) find (x-2y) / (x + 2Y)

Known: 3x2-xy-4y2 = 0 (x ≠ y) find (x-2y) / (x + 2Y) Known: the square of 3x - the square of xy-4y = 0 (x ≠ y) find (x-2y) / (x + 2Y)

It is known that 3x2-xy-4y2 = 0 (x ≠ y), and the deformation can be obtained
（X+Y）(3X-4Y)=0
Classification discussion:
When (x + y) = 0, then x = - y is substituted into the unknown value
Then (x-2y) / (x + 2Y) = - 3
When (3x-4y) = 0, then x = 4Y / 3 is substituted into the unknown value
Then (x-2y) / (x + 2Y) = - 1 / 5
Too long do not do so fine junior high school mathematics, oral arithmetic is not good, but the idea is also right, ha ha ~ ~ Gaogao first verify it! Help me score ha ha

(xy-x^2)-(x2-2xy+y^2)/xy×(x^2-y^2)/x^2 -xy-y^2

(xy-x^2) ÷ (x2-2xy+y^2)/xy×(x^2-y^2)/x^2
=-x(x-y)×xy/(x-y)²×(x+y)(x-y)/x²
=-y(x+y)
=-xy-y²

If 1A = 2B, 1b = 3C, then several a are equal to 36 C
Dear friends, pay attention! I not only want the answer, but also the detailed process!

1a=2b,1b=3c
The results are as follows: 1A = 2B, 2b = 6C
So: 1A = 6C
6a=36c
Six a equals 36 C

If x ＞ 0, y ＞ 0, and √ x (√ x + √ y) = 3 √ y (√ x + 5 √ y), find the value of (2x + 2 √ XY + 3Y) / (x - √ XY + y)
If x ＞ 0, y ＞ 0, and √ x (√ x + √ y) = 3 √ y (√ x + 5 √ y), find the value of (2x + 2 √ XY + 3Y) / (x - √ XY + y)

√x(√x+√y)=3√y（√x+5√y）
x+√xy = 3√xy+15y
x-2√xy-15y = 0
(√x+3√y)(√x-5√y) = 0
∵√x+3√y>0 ∴√x= 5√y
t=x/y = 25
(2x+2√xy+3y）/（x-√xy+y)
= (2t+2√t+3) / (t-√t+1)
= 38/21

1. (x-1) (X-2) (x-3) (x-4) - 24
2.2ax-10ay+5by-bx
3.ab(x+y)(x-y)-xy(a+b)(a-b)
4.9-a^2-6ab-9b^2
5.a^2-b^2-x^2+y^2-2ay+2bx

Point ABC is three points which are not on the same straight line in the plane, and point D is any point in the plane. If ABCD can form a parallelogram, it conforms to this rule in the plane
Point ABC is three points in the plane that are not on the same line, and point D is any point in the plane. If four points of ABCD can form a parallelogram, how many points D in the plane meet such conditions

There are three, because there are three points a, B and C that can be relative to d
I hope it can help you

Solve the equation 3x & # 178; - 12 = 0

3x²-12=0
3x²=12
x²=4
x=±2

It is known that the vertex of the parabola is at the origin, the axis of symmetry is the X axis, and the point m (- 2, m) on the parabola goes to the focus
It is known that the vertex of the parabola is at the origin, the axis of symmetry is the X axis, and the distance from the point m (- 2, m) on the parabola to the focus is equal to 6. Find the equation of the parabola and the value of M
Can you be more complete

The parabolic equation can be set as follows:
y²=2px,(p＜0)
According to the problem set and the definition of parabola
2 + | P / 2 | = 6 and M & # 178; = - 4P (P < 0)
The solution is as follows
p=-8.m=±4√2
Parabolic equation: Y & # 178; = - 16x
m=±4√2

It's urgent! Use arithmetic,
The students of three classes in grade five planted a row of 80 trees on the riverbank, counting from left to right, from the 58th to the right; from right to left, from the 63rd to the left, from the third class; so how many trees did class two plant?

80-58 = 22 (trees)
80-63 = 17 (trees)
80-22-17 = 41 (trees)
A: then class two planted 41 trees

As for the volt ampere characteristic curve of the small bulb, the high school physics experiment requires the ammeter to be connected externally, because the resistance of the small bulb is small, but isn't this the partial crimping method? The partial crimping method is suitable for measuring the resistance with larger resistance value, which I don't understand,

The sentence "partial crimping method is suitable for measuring the resistance with larger resistance value" only says that when the resistance to be measured is not much less than the resistance value of sliding rheostat, the current limiting connection method can not achieve the effect of greatly changing the voltage of the measuring circuit, so the partial crimping method has to be used. But it does not mean that the partial crimping method cannot be used for measuring the smaller resistance value, Third, the external connection of the ammeter in this experiment is because the internal resistance of the ammeter is not too small compared with the resistance of the bulb. If the internal connection method is used, the partial voltage of the ammeter can not be ignored, resulting in too large deviation in voltage measurement