Find (2x + Z-Y) / (x ^ 2-xy + xz-yz) - (2x + y + Z) / (x ^ 2 + XY + XZ + YZ)

Find (2x + Z-Y) / (x ^ 2-xy + xz-yz) - (2x + y + Z) / (x ^ 2 + XY + XZ + YZ)

The original formula = (X-Y) + (x + Z) / (X-Y) * (x + Z) - (x + y) + (x + Z) / (x + y) * (x + Z)
=1/(X+Z)+1/(X-Y)-1/(X+Z)-1/(X+Y)
=2Y/X^2-Y^2

(2X+Z-Y)/(X^2-XY+XZ-YZ)-(Y-Z)/(X^2-XY-XZ+YZ)

The answer is: (2 * x) / ((x - z) * (x + Z))

If three numbers x, y, Z satisfy XY / x + y = - 2, YX / y + x = 4 / 3, ZX / Z + x = - 4 / 3, then XYZ / XY + XZ + YZ=

It should be YZ / (y + Z) = 4 / 3 (y + y + Z) = 4 / 3xx / (x + y + y) = - 2 (x + y) / (x + y) / (x + y) / (XY) = - 1 / 21 / X + 1 / y = - 1 / 2 (1) YZ / (y + Z) = 4 / 3 (y + Z) / (YZ) = 3 / 41 / y + 1 / z = 3 / 4 (2) ZX / (Z + x) = - 4 / 3 (Z + x) / (ZX) = - 3 / 41 / x + 1 / z = - 3 / 3 / 4 (3) (3) (1) + (2) + (2) 2 (3) 2 (1 / x x x) 2 (1) 2 (1 / x x x x) 2 (3) 2) 2 (1 / x x x+ 1 / y + 1 / z) =

Given that three numbers x, y, Z satisfy XY / x + y = - 2. YZ / y + 4 = 4 / 3, ZX / Z + x = - 4 / 3, then XYZ / XY + XZ + YZ =?

(x + y) / (x + y) = - 2 (x + y) / (x + y) = - 1 / 2,1 / x + 1 / y = - 1 / 2yz / (y + Z) = 4 / 3, (y + Z) / (YZ) = 3 / 4,1 / y + 1 / z = 3 / 4, ZX / (Z + x) = - 4 / 3, (Z + x) / (ZX) = - 3 / 4 1 / Z + 1 / x = - 3 / 4 / 4 add the above three formula, get: 2 (1 / x + 1 / y + 1 / 1 / Z + 1 / z = 1 / 1 / 2 + 3 / 4 = 1 / 2 + 3 / 4 / 4 (1 / x + 1 / y + 1 / 1 / 1 / 1 / 1 / 1 / 1 / 1 / 4 / 4 / 4 / 4 / 4 z = - 1 / 4: (XY

Let x, y, Z satisfy XY x+y=-2，yz y+z=4 3，zx z+x=-4 3. Find XYZ XY + YZ + ZX

∵xy
x+y=-2，yz
y+z=4
3，zx
z+x=-4
3，
∴1
X+1
y=-1
2，1
Y+1
Z=3
4，1
Z+1
x=-3
4，
∴2（1
X+1
Y+1
z）=-1
2, i.e. 1
X+1
Y+1
z=-1
4，
Then XYZ
xy+yz+zx=1
One
X+1
Y+1
z=-4．

Given that three numbers x, y, Z satisfy XY / x + y = - 2, YZ / y + Z = 3 / 4, ZX / Z + x = - 3 / 4, then XYZ / XY + XZ + YZ =

If you take the reciprocal, you will get 1 / x + 1 / 1 / y = - 1 / 2 (1 / x + 1 / y = - 1 / 2) YZ / y + Z + Z = 3 / 4, then you get 1 / y + 1 / 1 / z = 4 / 3, ② ZX / Z + X + x = - 3 / 4, if you take the reciprocal, you will get 1 / x + 1 / z = - 4 / 3, 3 (1 / x + 1 / y + 1 / 1 / Z + 1 / Z + 1 / 2 + 4 / 3-4 / 3, then you get 1 / x + 1 / y + 1 + 1 / z = 1 / 4 / 4, the same division is (YZ + XZ + XY) / XYZ Z Z / Z / XYZ = XYZ + XYZ + XYZ + XYZ + XYZ + XYZ + xy= - 1 / 4 two

x. Y, Z are real numbers, and XY / x + y = 1 / 3, YZ / y + Z = 1 / 4, XZ / x + Z = 1 / 5, find the value of XYZ / XY + YZ + ZX

By inverting the denominator of XY / x + y = 1 / 3, YZ / y + Z = 1 / 4, XZ / x + Z = 1 / 5, the following equation can be obtained:
(1/x)+(1/y)=3 ①
(1/y)+(1/z)=4 ②
(1/z)+(1/x)=5 ③
Add (1 / x) + (1 / y) + (1 / z) = 6 (4)
We can get XY + YZ + ZX / XYZ = 6 if we divide 4 into four parts
So XYZ / XY + YZ + ZX = 1 / 6

Remove brackets and merge: (1) 3x ^ 3 - [x ^ 3 + (6x ^ 2-7x)] - 2 (x ^ 3-3x ^ 2-4x) (2)(3x^2-4y^2)+[-(x^2-2xy-y^2)]-[-(3x-2xy-y^2)]

(1)3x^3-[x^3+(6x^2-7x)]-2(x^3-3x^2-4x)=3x^3-x^3-6x^2+7x-2x^3+6x^2+8x=3x^3-x^3-2x^3-6x^2+6x^2+7x+8x=15x(2)(3x^2-4y^2)+[-(x^2-2xy-y^2)]-[-(3x-2xy-y^2)]=3x^2-4y^2-x^2+2xy+y^2+3x-2xy-y^2=3x^2-x^2-4y^2+y^2...

How to write 3x's Square - [7x - (4x-3) - 2x's Square] needs process, thank you

The square of 3x - [7x - (4x-3) - 2x squared]
=3x²-(7x-4x+3-2x²)
=3x²-7x+4x-3+2x²
=5x²-3x-3

(3x ^ 2) ^ 3-7x ^ 3 [x ^ 3-x (4x ^ 2 + 1] + (- x ^ 2) ^ 2, where x = 1 / 2 I calculate it is 5 / 4, but it is wrong, but I put 1 / 2 generation in, calculate out is 5 / 4, Baidu knows it is 87 / 68, I am very tangled, what is it?

(3x^2)^3-7x^3[x^3-x(4x^2+1]+(-x^2)^2
=27x^6-7x^3[x^3-4x^3-x]+x^4
=27x^6+21x^6+7x^4+x^4
=48x^6+8x^4
=48/64+1/2
=3/4+1/2
=5/4
You did the right thing!