Ask some eighth grade math fractional equation questions I have a question in my book, which is (100 / x) = (30 / X-7) Multiply both sides of the equation by X (X-7) at the same time, and remove the denominator to get 100 (X-7) = 30x If we solve this integral equation, we get x = 10 Test: substituting x = 10 into x (X-7), we get 10 × (10-7) ≠ 0 Just tell me in this way! 1.(x+1/x-1)-(4/x²-1)=1 2.(4x/2x-1)=(2x+1/x-2) 3.(7/x²+x)-(3/x-x²)=1+(7-x²/x²-1) 4.(x-2/x+1)-(16/x²-4)=1 5.(2-x/3+x)-(1/2)=(1/x+3)

1. (x + 1 / x-1) - (4 / X & # 178; - 1) = 1. Multiply both sides of the equation by (x + 1) (x-1) at the same time and reduce the denominator to get (x + 1) &# 178; - 4 = (x + 1) (x-1). To solve this integral equation, we get x = 1. Test: substituting x = 1 into (x + 1) (x-1), we get (1 + 1) (1-1) = 0 x = 1 is not the solution of the equation, so there is no solution 2. (4X / 2x-1) = (2x + 1 / X-2

1/[x(x-1)]+1/[x(x+1)]+… +1/[(x+9)(x+10)]=1/(x+10)

1/[x(x-1)]= 1/(X-1) - 1/X
1/[x(x+1)]=1/X -1/(X+1)
1/(X-1)-1/X+1/X-1/(X+1)+...+1/(X+9)-1/(X+10)
=1/(x+10)
1/(X-1)-1/(X+10)=1/(x+10)
1/(X-1)=2/(x+10)
2(x-1)=x+10
2x-2=x+10
x=12

In a factory, the working efficiency of group A is 20% higher than that of group B when two teams process the same kind of parts. Therefore, the time for group A to process 210 parts is half an hour less than that for group B to process 200 parts?

If group B processes x parts per hour, group A processes 1.2x parts per hour
210/(1.2X)=200/X -1/2
X=50
1.2X=60

Party A and Party B start from the destination of 6 km and 10 km respectively. The speed ratio of Party A and Party B is 3:4. Party A arrives 20 minutes earlier than Party B. ask for Party A,

Let a's speed be 3x and B's speed be 4x (according to the condition: the speed ratio of a and B is 3:4)
20min=1/3h
So 10 / 4x-6 / 3x = 1 / 3
Multiply both sides by 12x at the same time
30-24=4x
6=4x
x=1.5
So the speed of a is 3 * 1.5 = 4.5km/h
The speed of B is 4 * 1.5 = 6km / h
A: the speed of a is 4.5 km / h, and that of B is 6 km / h

If the value of fraction x + 2 / X-1 is an integer, find the value range of X
If the value of fraction x + 2 / X-1 is an integer, find the integer value of X

Let x + 2 / X-1 = n (n is an integer)
The reduction is: x = 1 + 3 / (n-1)
If x is an integer, then n = - 2,0,2,4
That is, x = 0, - 2,4,2
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If the fraction 2x-1 is divided by x ^ 2 + 5x-6, the value range of X is

If the fraction 2x-1 is divided by x ^ 2 + 5x-6, the value range of meaningful x is
x^2+5x－6≠0
(x+6)(x-1)≠0
X ≠ - 6 or X ≠ 1

According to the solution of fractional equation 9 / 4-x square = 2 / 2 + X

9 / 4-x square = 2 / 2 + X
9/(2+x)(2-x)=2/(2+x)
9=2(2-x)
9=4-2x
x=-2.5

Evaluation after simplification: (a + 3 / A ^ 2-4 + A / 6-a-a ^ 2) divided by 4A + 9 / 5a-10, where x = 8

[(a + 3) / (a + 2) (A-2) - A / (a + 3) (A-2)] × 5 (A-2) / (4a + 9) = [(a + 3) / (a + 2) - A / (a + 3)] × 5 / (4A + 9) = [(a + 3) & # 178; - A (a + 2)] / (a + 2) (a + 3) × 5 / (4a + 9) = (4a + 9) / (a + 2) (a + 3) × 5 / (4a + 9) = 5 / (a + 2) (a + 3) when a = 8, the original formula = 5 / 10 × 11 = 1 / 22

If 3 is less than X and less than 8, simplify: | X-9 | + | X-2|=

The solution equation is: (1) 2 / x = 3 / x + 1 (2) 1 / x-4 = 8 / x ^ 2-16, simplifying: (12 / m ^ 2-9) + (2 / 3-m)

（1）2/x=3/(x+1)
2(x+1)=3x
2x+2=3x
x=2
(2)1/(x-4)=8/(x^2-16)
8(x-4)=x^2-16=(x+4)(x-4)
(x-4)(x+4-8)=0
(x-4)^2=0
x-4=0
x=4
In the test, x = 4 makes x-4 and x ^ 2-16 equal to 0
Not established
So the original equation has no solution
12/(m^2-9)+2/(3-m)
=12/(m^2-9)-2/(m-3)
=12/(m^2-9)-2(m+3)/(m-3)(m+3)
=(12-2m-6)/(m-3)(m+3)
=(6-2m)/(m-3)(m+3)
=2(m-3)/(m-3)(m+3)
=2/(m+3)