Asked about the math problem of the first year of high school, please come in, thank you

1. The meaning of the title is expressed in the form of a solution set, and the definition of the set is:

In addition, a single element set can be represented in this form.

If you use , then, it means: as long as x 1, no matter how many y you are, it is the solution of the equation; As long as y 2, no matter how many x there are, it is also the solution of the equation, which is obviously not true.

If there is an x equal to a number, there is only one y 3-x corresponding to this x, so (x,y) denotes a pair of real numbers, so the solution set should be represented by or .

2. Simplification obtains: (m-1)2 (2 means squared) > means greater than or equal to) 1 gets: m-1>=1 or m-1<=-1 can get: m 2, or m 0 squares a number, you can get positive and negative two solutions.

The first x-y=-1 y=2 doesn't solve the equationSo no.

x=1,y=2} should be written as a set. This is a misspelling.

The second 4 (m-1) 2 - 4 0

So there is (m-1) 2 1

So there's m-1 1 or m-1 -1

So the answer is m 2, or m 0

The first one doesn't understand what you mean.

2nd: (m-1) square 1

M-1 1 or M-1 -1

The solution is m 2, or m 0

I didn't understand the first course, but the second one seems to be a junior high school student.

Because 4(m-1)2-4 is greater than or equal to 0

Therefore, (m-1)2 is greater than or equal to 1

When m-1 is greater than 0, m-1 is greater than or equal to 1

So m is greater than or equal to 2

When m-1 is less than 0, m-1 is less than or equal to -1

So m is less than or equal to 0

I didn't see it in the first question.

Question 2 4 (M-1) 2 - 4 0

4(m-1)2 ≥4

m-1)2≥1

M-1 1 or M-1 -1

m 2, or m 0

The deformation of the equation is: x 2-x + ab + b = 0

Because there is only one element.

So there is only one solution to the equation, and x=a

So discriminant = 0

Substituting x=a into the equation and calculating the discriminant.

1-4（ab+b）=0

a^2+ab+b-a=0

Because: 4(ab+b)=1, so ab+b=1 4, so: a 2+1 4-a=0 i.e. (a-1 2) 2=0 solution: a=1 2

So: b=1 6

There is only one solution, and there is a discriminant formula to obtain 1-4*(ab+b)=0 a is the solution of the original formula, and the |a^2+ab+b=a a=-1/2 b=1/2

Root number (1+a) 1 root number (1-b) = root number (1+a)(1-b) = root number 1+a-b-ab

1 b-1 a>1 =>a-b-ab>0 so the root number 1+a-b-ab>1 , so the root number (1+a) is greater than 1 root number (1-b).

Certificate: Connect DF Nexus AE to DC to O'The connection de is because d f is the midpoint of the edge.

According to the median line theorem or direct proof that ADE is similar to ABC, DF is parallel and equal to 1 2BC

So DOF is similar to COB

So bo:of=co:od=bc:fd=2, because d e is the midpoint of the edge.

The same goes for AO'C vs. EO'd in ao':o'e=co':o'd = 2 because c o o'd four-point collinear and co:od=co':o'd=2 so point o'Coincident at point O

So both questions are proven.

Level 1 users can't illustrate, just look at the 4th floor.

(1) Bring x=1 in, and get 1 a+b+c 1 from the meaning of the question, so a+b+c=1

Bring ab c 1 into the inequality 3ab + 3bc + 3ac a b c 2ac 2bc 2ab

2ab+2bc+2ac 2a 2b 2c , from the fundamental inequality a b 2ac, the same can be obtained ......(Don't omit it).

2) Bring x=-1 into a-b+c=0 to get b=1 2, bring in the original formula, the axis of symmetry of f(x) -b 2a=-1 4a, because f-1) f(1), so -1 4a 0 or -1 4a -1

The solution of a 0 parabola is open downward, which does not conform to x f(x) rounding, the solution is a 1 4, and x=0 is brought into the solution c 1 4

So a=1 4, c=1 4, b=1 2, so the expression f(x) is ......(Write it yourself).

This can be done according to the area method.

Connect AE, AO, OE, DF

For FAB, df is the midline, S DBF=S ADF

For ADC, df is the midline and s def=s adf

s dfc=s dfc and dof are common parts s dob=s foc

For BOC OE for the midline, S OBE=S OEC

For AOB OD is the midline, S ODB=S ODA is the same as S Ofc=S Ofa

S odb=s ofc s oad=s oaf

S quadrilateral AOEB=S quadrilateral AOEC and for ABC, AE is the midline, S ABE=S AEC

AE coincides with AOE, i.e., AOE is three-point collinear.

In ABC, D,F is the midpoint of the median line DF DF BC and DF=1 2BC

DFO dbo of ob=do oc=df bc=1 2 The same goes for eo oa=1 2

Interesting: I haven't played geometry in a long time.

It's a little vague, but I looked at it, and this should be able to get the midline from the midpoint, and the midline can be equigonal, and the equiangular compared to the edge can be equilateral, in simple terms, through similar triangles.

What a mess, I didn't go to high school. What is the practical significance of these? (⊙

...Just use the center of gravity.

Sometimes don't overcomplicate it.

Mom? I was so dizzy to see it, I flashed.

Let's say the definition of the center of gravity, and then we can come to a conclusion.

Substitution is enough, the equation is substituted into a, its solution is 3 and 1, find a, b and then substitute back b