Elementary bipartite equations

The detailed solution is as follows:

The left side is divided = -x+5 x2-3x+2, and the right side is divided = -x+5 x2-7x+12, because the left side is equal to the right side, so -x+5 x2-3x+2=-x+5 x2-7x+12, so x=5 2

The left and right sides of the equation are respectively divided into [3(x-1)-4(x-2)] [(x-2)(x-1)]=[(x-3)-2(x-4)] [(x-4)(x-3)] to obtain (-x+5) [(x-2)(x-1)]=(-x+5) [(x-4)(x-3)], then -x+5=0, so x=5

5 or left = (5-x) (x-2)(x-1).

Right = (5-x) (x-4)(x-3).

The denominator is not 0, and the equation holds when the numerator is 0, i.e. x=5.

When the numerator is not zero, x=

First, left = (5-x) (x-2)*(x-1), right = (5-x) (x-4)*(x-3).

Just make the denominator equal to find x=

It's okay to get through first.

5-x) (x-2)(x-1) = (5-x) (x-4) (x-3) so. 5-x) (x-4) (x-3) = (5-x) (x-2) (x-1) x=5 or (x-4) (x-3) = (x-2) (x-1) x1 = 5, x2 = 5 2

/(x+4)-1/(x+7)=1/(x+3)-1/(x+6)

Shift -1 (x+6) and -1 (x+7) respectively, so that the values of 6+4 and 7+3 are all about 10, and the numerator is 2x+10, and the denominator is different, and the fractional value is the same, and only the numerator can be 0, that is, 2x+10=0, x=-5

2、(x-4)/(x-5)-(x-7)/(x-8)=(x-5)/(x+6)-(x-8)/(x-9)

In the same way as the first question, the direction is shifted, and the form of addition is changed, the molecule is deformed, x-5+1, x-9+1, x-8+1 on the right, x-6+1, guaranteed, 5+9 and 8+6 are both 14, and the 1 on the left and right is eliminated, that is, 2x-14=0, x=7

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

Move 1 to factorization, this question is not very well designed.

(x^2)-25]=3/[(x^2)+8x+15]+5/[(x^2)-2x-15]

Factoring the denominator, removing the denominator, becomes 6 (x+3) = 3 (x-5) + 5 (x+5) to get x=4

5、(x+2)/(x+1)-(x+4)/(x+3)=(x+6)/(x+7)-(x+8)/(x+7)

There's something wrong with that.

x+1)-1/x=1/(x-2)-1/(x-3)

The same is true (same as a question): 2x-2=0x=1

1/(x+6)(x+7)=1/(x+3)(x+4)x= -52.、(x-4)/(x-5)-(x-5)/(x+6)=(x-7)/(x-8)-(x-8)/(x-9)

Same as above. Move to the right of the equation and simplify.

At present, an average of x tons of coal are mined per day.

33000/x =23100 /(x-330)23100 x=33000（x-330）

x=1100

The original plan was to mine x tons of coal per day, but now the average daily coal mining x + 330 tons is 23100 x = 33000 (x + 330).

At present, an average of x tons of coal are mined per day.

33000/x =23100 /(x-330)23100 x=33000（x-330）

x=1100

The original plan of coal mining x tons per day is now an average of x + 330 tons per day.

23100/x=33000/(x+330)

Solution: Set the specified date as x days, and the total project as 1

Then the efficiency of A is 1 x and B is 1 (x+3).

1 x+1 (x+3))*2+(x-2)(1 (x+3))=1 solution. x=6

After setting more than 5 cubic meters, each cubic meter will be charged x yuan, and Xiao Wang's household water consumption is y cubic meters, and Xiao Li's family is 3y 2

Then the equation is 5*, 5*

The solution is: x=2, y=15, so the excess part is charged 2 yuan per cubic meter.

If the part exceeds 5 cubic meters, the charge is $x. Xiao Li's family's water consumption this month is y, and the highest charge when the water consumption does not exceed 5 cubic meters is 5* yuan, so the water consumption of both families exceeds 5 cubic meters.

2/3y-5）x+ （y-5）x+

y=15 x=2

The excess part is charged at 2 yuan per cubic meter.

So both of them are over 5 tons, and if the excess part of the unit is x yuan tons, then:

For the Zhang family, the cost of 5 tons is, so the cost of 5 tons is 10 x of the water consumption outside of 5 tons, so that the total water consumption is 5 + 10 x The same can find out the water consumption of the Li family, so that there is:

5+10/x):(5+20/x)=2:3

10+40/x=15+30/x

x=2, that is, the part exceeding 5 tons will be charged 2 yuan per ton.

Suppose Xiao Li's household water consumption is a

The final answer is 15 yuan cubic meters.

Set the excess part of the charge to be x yuan, and Xiao Li's household water consumption is ym xy=20+5x2 3xy-5x=10 will be brought in.

40/3+10/3x-5x=10

The solution is x=2

1/(x-1)(x-2)=1/（x-2) -1/(x-1)1/(x-2)(x-3)=1/（x-3) -1/(x-2)..

1/(x-2010)(x-2011)=1/（x-2011) -1/(x-2010)

The addition on the right side of the equation cancels out the same term, which is equal to 1 (x-2011)-1 (x-1).

(3x-1) (x+1) = (6x+1) (2x+1) multiplied by (x+1) (2x+1).

3x-1)(2x+1)=(x+1)(6x+1)6x²+x-1=6x²+7x+1

6x=-2x=-1/3

Fractional equations to be tested.

Upon examination, x=-1 3 is the root of the equation.

x1 is the root seven minus two.

x2 is the negative root seven minus two.

This score is derived from the question (x+5) x

The equation is obtained according to the second known condition.

x+5)-1] (x+14)=x (x+5), i.e. (x+4) (x+14)=x (x+5) gives x=4

The score is 9 to 4