(x + 1 / 3 x minus X-1 / x) divided by the square of x minus 1 / 2 will help to calculate the following process

(x + 1 / 3 x minus X-1 / x) divided by the square of x minus 1 / 2 will help to calculate the following process

[3x/(x+1)-x/(x-1)]÷(x-2)/(x^2-1)
=(2x^2-4x)/(x^2-1)÷(x-2)/(x^2-1)
=2x(x-2)/(x-2)
=2x

The addition and subtraction of fractions [x-5-x-2] divided by [3 X-6 3-x]

=[5/(x-2)-(x+2)]×[-3(x-2)/(x-3)]
=(5-x²+4)/(x-2) ×[-3(x-2)/(x-3)]
=-(x+3)(x-3)/(x-2)×[-3(x-2)/(x-3)]
=3(x+3)
=3x+9

The solution of the fractional equation x squared plus 5 / x minus x squared minus 1 / x equals 0

That is, 5 / (x? + x) = 1 / (x? - x)
5x²-5x=x²+x
4x²-6x=0
x(2x-3)=0
ν x = 0 (root augmentation) x = 3 / 2

A fractional equation x squared plus x divided by 7 plus x squared minus 3 equals x squared minus 1 divided by 6

From the meaning of the title, we can see that x ≠ 0 and X ≠ 1 and X ≠ - 1
Both sides of the original equation are multiplied by X (x + 1) (x-1)
7(x-1) + 3(x+1)=6x
7x-7+3x+3=6x
10x-4=6x
4x=4
The solution is: x = 1
It is easy to know that x = 1 is the augmented root of the original equation
So the original equation has no real number solution

Solving the fractional equation: the difference between 3x-6 divided by X squared minus 4 is 1 divided by X + 2

(3x-6)/(x²-4)=1/(x+2)
Both sides of the equation are multiplied by x 2 - 4, i.e. (x + 2) (X-2)
3x-6=x-2
2x=4
X=2
It is proved that x = 2 is an augmented root, so the original equation has no solution

X plus one third minus three minus two parts of X is the square of x minus 12 of nine

That is, 1 / (x + 3) + 2 / (x-3) = 12 / (x + 3) (x-3)
Double (x + 3) (x-3)
x-3+2x+6=12
3x=9
X=3
After testing, when x = 3, the denominator x-3 = 0
It's root augmentation
Give up
So the equation has no solution

When x=_____ The fraction 2x-1 molecule X-6 is meaningless=______ The fractional value is zero

When x=__ One in two___ The fraction 2x-1 molecule X-6 is meaningless=___ 6___ The fractional value is zero

Xiaoming asked Xiaohong, when x is the value of X, the fraction X / x ^ 2 + 2x is meaningless? Xiaohong answer: because X / x ^ 2 + 2x = 1 / x + 2, x + 2 = 0, x = - 2, so when x = - 2, the fraction is meaningless. Question: is there any error in Xiaohong's answer? If so, please help Xiaohong find out the cause of the error and correct it

There is a mistake
To make a fraction meaningful, the denominator must not be 0. The denominator is x ^ 2 + 2x. When x ^ 2 + 2x = 0, x = 0 or x = - 2. So when x = 0 or x = - 2, the fraction is meaningless
Can't divide directly

When x = - 2, X-B of fraction x-a is meaningless; when x = 4, the value of X-B in fraction x-a is 0, then what is a + B

The fraction (X-B) / (x-a) is meaningless when x = a, then a = - 2
In the same way, when x = B, the value of this fraction is 0, we get: B = 4
Then: a + B = (- 2) + 4 = 2

When x = 1, the fraction is meaningless; when y = - 1, the value of fraction is 0. Try to find the value of a? + B

When y = - 1, the value of fraction is 0, so (2x + a) / (y-b) = 0, y = b = - 1, so x = 1, put (2x + a) / (y-b) = 0, y = b = - 1, so x = 1, put (2x + a) / (y-b) = 0, deform 2x + a = (y-b) * 0, replace x = 1, y = b = - 1 to get a = - 2, B = - 1, a = -