Junior 2 math fractional equation problems, speed

The question is: 1 "(x-2)*(x-3)" + 1 "(x-1)*(x-2)" = 1 (x-1)?

The answer is: 5 The process is troublesome, and I can't understand it.

1/《（x-2）*（x-3）》+1/《（x-1）*（x-2）》-1/（x-1）=0

1/《（x-2）*（x-3）》+1/（x-1）*《1/（x-2）-1》=0

1/《（x-2）*（x-3）》+1/（x-1）*《1/（x-2）-（x-3）/(x-3)》=0

1/《（x-2）*（x-3）》+1/（x-1）*《x-3-x^2+5x-6)/(x-2)*(x-3)》=0

1/《（x-2）*（x-3）》-1/（x-1）*《x-3）^2/(x-2)*(x-3)》=0

1/《（x-2）*（x-3）》-x-3)/《（x-1）*(x-2)》=0

1/《（x-2）*（x-3）》=(x-3)/《（x-2）*（x-1）》

1/《（x-2）*（x-3）》=(x-3)/《（x-2）*（x-1）》

1/（x-3）=(x-3)/（x-1）

x-3)^2=x-1

x-2)(x-5)=0

x can't be equal to 2, why, you should know.

So x=5

Solution: Original formula = 1 (x-2)-1 (x-3) 1 (x-1)-1 (x-2)=1 (x-1) gives 1 (x-3)=0, because the numerator is 1, so the value cannot be 0, so the equation has no solution.

1/（ （x-2）*（x-3））=1/（x-3 ） 1/（x-2）

1/（ （x-1）*（x-2） ）=1/（x-2） -1/（x-1）

So the left side of the equation is equal to 1 (x-3) -1 (x-1).

Untie; There are x people in the second middle school. Then there are 2x-10 people in middle school A who are solved by 7600 (2x-10)=4000 x and x=100

So 2x-10=190

Solution: Let B process x per hour, then A processes 2 and 1 2*x per hour.

1500 (2 again, 1 2*x)+18=1500 x600 x+18=1500 x

Multiply x at the same time to get:

600+18x=1500

18x=900

x=50 root test: After testing, x=50 is the root of the original equation (the root test of the fractional equation is essential) 2 and 1 2*50=125

A: A processes 125 pieces per hour, and B processes 50 pieces per hour.

Solution: Let the bicycle speed be x kilometers per hour and the car speed be kilometers per hour.

20 x + multiply x at the same time to get :

20+x=12 is examined, and x=12 is the root of the original equation.

A: Bicycles travel 12 kilometers per hour and cars travel 30 kilometers per hour.

Let A process x per hour, then one will process 2x 51500 x + 18 = 1500 per hour

x=125 B 50

Let the bicycle x then the car 2x

50/x=50/2x+

Get x=10 for cars 20 for bikes 10

Set x number of B. A is two and a half times the size of B.

So A is one.

Time A is 1500 hours, B is 1500 x hours.

B has 18 hours longer than A.

So 1500 x-1500

1500/x-600/x=15

900/x=15

x = 150 for A, 60 for B.

The ratio of A to B is 4:5

Set 4x days for A and 5x days for B.

A did 4 days and completed 4 (4x), and B did 4+16 days and completed (4+16) (5x).

So 4 (4x)+(4+16) (5x)=11 x+4 x=1

5 x = 1 so x = 5, 4 x = 20, 5x = 25 so A 20 days, B 25 days.

1.If A produces x, then B produces 2 5x parts per hour.

1500 (2 5 x)-1500 x=18 solution equation x=75 is obtained

So A produces 75 per hour and B produces 50 per hour.

2.If the speed of the bicycle is x, then the speed of the car is.

Draws 50+

Solve equation x=12

So the speed of a bicycle is 12 kilometers per hour, and the speed of a car is 20 kilometers per hour.

One set B processes x parts per hour.

Two and a half = 5 2

1500 x 1500 (5 2 times x) 18x=50 A50 5 2=125

The speed of the bicycle is set to x kilometers per hour.

50 x 12 cars.

x/(x-2)-(x-2)/(x-2)=8/[(x+2)(x-2)]x-(x-2)=8/(x+2)

2=8/(x+2)

2(x+2)=8

2x+4=8

x=2 because the original denominator x-2=0

x=2 is the root addition of the original equation, and the original equation has no solution.

Solution: The original equation can be reduced to x(x+2)-x +4-8=0

The solution of x=2 is tested to be the increasing root, so there is no solution to this equation.

(1) Multiply both sides of the equation by 3x at the same time to get: 2(x 2) -6x -3=0 so x= (3+ 15) 2 or x=(3- 15) 2

2) 2 x=-1 i.e. x= -2

3) x 2 +1 2+1=6x+6 5 i.e. (11x) 2 =3 10 so x=3 55

Sort out the above equation to get 2x 3 6x 0, and get x= (3+ 15) 2 or x=(3- 15) 2

Finishing, -1=2 x, x=-2

(1) (4x) 3 - 2 x - 4=0 i.e. 2(x 2) -6x -3=0

So x= (3+ 15) 2 or x=(3- 15) 2(2) 2 x=-1 i.e. x= -2

3) x 2 +1 2+1=6x+6 5 i.e. (11x) 2 =3 10

So x=3 55

(1) (4x) 3 - 2 x - 4=0 i.e. 2(x 2) -6x -3=0

So x= (3+ 15) 2

or x=(3- 15) 2

2) 2/x=-1

i.e. x= -2

3) x/2 +1/2+1=6x+6/5

i.e. (11x) 2 =3 10

So x=3 55

Assuming that A completes a every day, B completes B every day, and the construction period is x days, then 3a + b * x = a * x = b * x + 6b, 3a = 6b, a = 2b, substitute into the original formula, 2 * b * x = b * x + 6b, x = 6, scheme 1: 72000, scheme 2: delay, scheme 3: 66000, so choose scheme 3.

If A needs x days to dry alone, then B needs x + 6 days to dry alone, A needs x to dry every day, and B needs to do 1 (x+6) per day

According to column (3): 3(1 x+1 (x+6))+x+3) (x+6)=1, the solution is: x=6

1) Project payment: 6*

2) Project payment: (6+6)*

3) Project payment: 3*(

So option 2 is the most economical.

Let x2=y, then the original equation is y+1 y=2+2a, so.

y=2+2a, i.e. x2=2+2a, because a -1, so 2+2a 0, so x= 2+2a