How to solve this math problem How to solve this math problem?

From the question, we can get 3x+, that is, remove the middle formula) and then simplify: 3x-x=1

That is: 2x=1

Divide both sides of the equation by 2 at the same time to get x=1

Then remove the first formula in the original question to get:

2x+y/5=x-y+1/6

Bring x=1 into the :

That is:1

Move y to the same side of the equation to get :

y 5 + that is: 9y 20 = 1 12

Solution: y=5 27

Then x=1 I'm from Shandong, it's different from your book, I've written the most detailed, I hope you can understand it.

Solution: The formula 3x+2y 4 = 2x+y 5 = x-y+1 6 can be reduced to:

3x+2y\4 = 2x+y\5

2x+y\5= x-y+1\6

This system of equations is reduced to 60x+10y=40x+4y(1)60x+6y=30x-30y+5(2).

From (1) we get x=-3 10y(3).

From (2) we get 30x=-36y+5(4).

Substituting (3) into (4) gives y=5 27

Substituting y=5 27 into (3) gives x=-1 18

So the solution of this system of equations is x= -1 18

y= 5/27

The above equation can be simplified as:

3x+y 2=2x+y 5=x-y+1 6 The following system of equations can be obtained by the above equation.

The first: 3x+y 2=2x+y 5, simplified: x+3y 10=0, the second: 2x+y 5=x-y+1 6, simplified: x+6y5=1 6 The solution of the above equations gives x=-1 14, y=5 27

Simplify: 180x+30y=120x+12y=60x-60y+10 and then turn it into a system of binary linear equations.

180x+30y=120x+12ySimplified: 60x+18y=0120x+12y=60x-60y+10 Simplification: 60x+72y=10 Solutiony=5 27,x=-1 18

This is a problem for solving binary equations.

Solution: According to the meaning of the topic:

3x+2y/4=2x+y/5---1

2x+y/5=x-y+1/6---2

From (1) we get x=-3 10y(3).

Replace 3 with 2.

y=5/27

Replace y=5 27 into 3.

x=-1/18

I don't have a good math head

But it can definitely be done.

Maybe the way to compare benzene.

Do you mean to divide?

3x+2y) times 5 times 6=(2+y) times 4 times 6=(x-y+1) times 4 times 5 will the rest be solved?

I don't know which version, but this can be solved with a continuous equation.

Sweaty, difficult, the questions in junior high school are so difficult, and I will suffer in junior high school in the future.

It's too hard for students right now.

Original = (2019x2020+2019) 2020x2019+2021

In 2022, I raised my hand again, and I was imitating the potato tung 2020

Original style trace hall = (）×1/2019 + 2021 ..Rewrite with fractions.

×1/2019+2021 ..The multiplication of the posture is faintly fingered.

= 2022 and 1 2020

log3[27^(1/4)/3]

log3[(3³)^1/4)/3]

log3[3^(3/4-1)]

log3[3^(-1/4)]

-1/4)log3(3)

log25+log4 [What's the end?] Let's say it's 10]LG25+LG4

lg(25*4)

lg100lg10²2lg10

27^log7(2)

2, [You can think of it like this: let x=7 log7(2), take the logarithm with 7 as the base, log7(x)=log7[7 log7(2)], =log7(2)*log7(7), log7(2)*1, log7(2), x=2].

It is sufficient to find a derivative of x on both sides of the equation at the same time.

If the division is greater than zero, it can be discussed on a case-by-case basis. Both up and down are greater than 0; Or the top and bottom are less than 0, that is, when the same positive and the same negative, it is greater than or equal to 0

For example, in the first question, when the top and bottom are greater than or equal to 0, it must be greater than or equal to -1 and greater than or equal to -2, then the satisfactory answer is greater than or equal to -1

When the top and bottom are less than or equal to 0, less than or equal to -1 and less than or equal to -2, then the answer that satisfies is less than or equal to -2

A, x is greater than or equal to -1 or less than or equal to -2

If it is less than one positive and one negative, the upper is greater than 0 and the lower is less than 0, and a solution is obtained: the lower is greater than 0 and the upper is less than 0, and a solution is obtained (provided that this solution is true.) For example, there is no number greater than 0 and less than -2).

Let's take the first question for example.

x+1)/(x+2) >= 0

Equivalent to (x+1)*(x+2) >= 0 and x+2 is not equal to 0, the same is true for others.

Let the number in the lower right corner be x, then there is:

11-x) -2-x) = 9(14+x) -x =1425 2 does not fit the meaning of the question and cannot be solved.

a-b=9 (1)

c-d=14 (2)

a+c=12 (3)

b+d=2 (4)

1) +2) to get a-b+c-d=23

i.e. (a+c)-(b+d)=23

However, from (3) and (4), it can be seen that (a+c)-(b+d)=1023≠10 has no solution.

Let the number in the lower right corner be x, then there is:

11-x） -2-x） =9

14+x） -x =14

Row mu = file sen sen =

25 Spring difference 2

There is no solution to the topic.

Because the two roots are opposite to each other, the sum of the two roots is 0.

x1+x2=-（m²-4m-5）=0

m²-4m-5=0

m-5）（m+1）=0

m-5 = 0 or m+1 = 0

So m=5 or -1

Substituting m=5 into the original equation yields.

x²+5=0

x =-5 is obviously rootless at this point, so m=5 (rounded).

Substituting m=-1 into the original equation yields.

x²-1=0

x = 1 x = 1 or -1

The sum of the two is exactly 0, which satisfies the topic.

In summary, m=-1

The difficulty of this problem is that the calculation of m is closed, and the reverse generation is forgotten to go back to verify and remove the added roots).