Solve the 20 fractional equation calculation problems, which are the process

1)2x+xx+3=1

Multiply both sides of the equation by x(3+3) and remove the denominator.

2(x+3)+x2=x2+3x, i.e. 2x 3x= 6

So x=6

Test: When x=6, x(x+3)=6(6+3)≠0, so x=6 is the root of the original fractional equation.

2)15x=2×15 x+12

Multiply both sides of the equation by x(x+12) and reduce the denominator.

15(x+12)=30x.

Solve this integer equation and get.

x=12.Test: When x=12, x(x+12)=12(12+12)≠0, so x=12 is the root of the original fractional equation.

3）2(1x+1x+3)+x－2x+3=1.

I tidy it up and get it. 2x+2x+3+x 2x+3=1, i.e., 2x+2+x 2 x+3=1, i.e. 2x+xx+3=1

Multiply both sides of the equation by x(x+3) and remove the denominator.

2(x+3)+x2=x(x+3), i.e. 2x+6+x2=x2+3x, i.e. 2x 3x= 6

Solve this integer equation and get x=6

Test: When x=6, x(x+3)=6(6+3)≠0, so x=6 is the root of the original fractional equation.

4）2x-3+1/(x-5)=x+2+1/(x-5)

Subtract 1 (x-5) from both sides at the same time to get x=5

Substituting the original equation so that the denominator is 0, so x=5 is the increment.

So there is no solution to the equation!

Test format: bring x=a into the simplest common denominator, if x=a makes the simplest common denominator 0, then a is the root of the original equation. If x=a makes the simplest common denominator not zero, then a is the root of the original equation.

5）x/(x+1)=2x/(3x+3)+1

Multiply both sides by 3 (x+1).

3x=2x+(3x+3)

3x=5x+3

2x=3 x=3/-2

After testing, x=-3 2 is the solution of the equation.

6）2/（x-1）=4/（x^2-1）

Multiply (x+1) (x-1).

2(x+1)=4

2x+2=4

2x=2 x=1

Substituting x=1 into the original equation, the denominator is 0, so x=1 is the increment.

So there is no solution to the original equation.

7）3x/1-x-1/x-1=1

Solution: Multiply both sides of the equation by (1-x) at the same time to obtain.

3x+1=1-x

x=0 test: x=0 is the solution of the original equation.

8）2/1+x-3/1-x=4/x^2-1

Solution: Multiply both sides of the equation by (x 2-1) at the same time.

2（x-1）+3（x+1）=4

x=3 5 tested: x=3 5 is the solution of the original equation.

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

3x=2x+(3x+3)

3x=5x+3

2x=-3x=-3/2

Test: x=-3 2 is the solution of the equation.

1+x-3 1-x=4 x 2-1 The equation is multiplied by (x 2-1) on both sides at the same time

2（x-1）+3（x+1）=4

x=3 5 test: x=3 5 is the solution of the original equation.

x-3+1 (x-5)=x+2+1 (x-5) subtract 1 (x-5) from both sides at the same time to get x=5

Test: Substituting the original equation so that the denominator is 0, so x=5 is the root increase, so the equation has no solution!

（x-1）=4/（x^2-1）

Multiply (x+1) (x-1).

2(x+1)=4

2x+2=4

2x=2x=1 test: substituting x=1 into the original equation, the denominator is 0, so x=1 is the root. So there is no solution to the original equation.

x/1-x-1/x-1=1

Both sides of the equation are multiplied by (1-x) at the same time

3x+1=1-x

x=0 test: x=0 is the solution of the original equation.

Look at this 8 (4x 2-1)+(2x+3) (1-2x)=1

8/(4x^2-1)-(2x+3)/(2x-1)=1

8/(4x^2-1)-(2x+3)(2x+1)/(2x-1)(2x+1)=1

8-(2x+3)(2x+1)]/(4x^2-1)=1

8-(4x^2+8x+3)=(4x^2-1)

8x^2+8x-6=0

4x^2+4x-3=0

2x+3)(2x-1)=0

x1=-3/2

x2=1/2

In the substitution test, x=1 2 makes the denominator 1-2x and 4x 2-1=0. Abandon it.

So the original equation solves: x=-3 2

x+1)/(x+2)+(x+6)/(x+7)=(x+2)/(x+3)+(x+5)/(x+6)

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

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

1/(x+2)+1/(x+7)=1/(x+3)+1/(x+6)

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

x+3-(x+2))/(x+2)(x+3)=(x+7-(x+6))/(x+6)(x+7)

1/(x+2)(x+3)=1/(x+6)(x+7)

x+2)(x+3)=(x+6)(x+7)

x^2+5x+6=x^2+13x+42

8x=-36

x=-9/2

Upon examination, x=-9 2 is the root of the equation. Reference.

There is a problem here, but you need to manipulate it to see it.

1.Click on the webpage.

2.From the "Knowledge Point Selection" on the left, find the fraction - fraction equation 3Find "Enter Question Basket", and next to it there is a Select Question Type Select Calculation Question point filter.

OK, there's a lot of it and the answers.

There must be no less than 50 courses.

Question 1: Question 1 y=(5x 68) 9 Question 2 According to the meaning of the question, the average score of 6-9 games = 68 4=17 Because x and y must be less than 17, the total score of the first five games cannot be greater than or equal to 17*5=85, so the maximum value.

.Click on the webpage.

2.From the "Knowledge Point Selection" on the left, find the fraction - fraction equation 3Find "Enter Question Basket", and next to it there is a Select Question Type Select Calculation Question point filter.

OK, there's a lot of it and the answers.

There must be no less than 50 courses.