Please teach me the topic of Mathematics Compulsory 5 in the second year of high school, thank you

Hello, the methods used for this kind of problem to find the fixed point are all separated parameter method.

This question separates m

2mx-x-my-3y-m+11=0

2x-y-1)m-x-3y+11=0

Let 2x-y-1=0 and -x-3y+11=0

We get x=2, y=3

So over the fixed point (2, 3).

Method 1: Special value method.

Since m can be taken as any value, then substituting m=0, m=1, we can get the straight line -x-3y+11=0, and x-4y+10=0, and find the intersection point of two straight lines to get the point (2,3), so this fixed point is (2,3).

Method 2: General method.

Replace the straight line with the form y=k(x-a)+b, that is, the straight line is constant over the fixed point (a, b), so the original straight line is y=[(2m-1) (m+3)](x-2)+3, and it is constant over the fixed point (2,3)

There are also some of my thoughts, if it is a solution question, you can first use the special value method to find the fixed point, and then use the general method to make it up when writing the process, so that you know which number to make up, because if you don't know the fixed point, some questions are difficult to make up that form.

Hope it helps.

2m-1)x-(m+3)y-(m-11)=0 to 2mx-my-m-x-3y+11=0

That is, m(2x-y-1)-(x+3y-11)=0 can be seen that regardless of m as any real number, when 2x-y-1 and x+3y-11 are 0 at the same time, that is, x=3, y=2, (2m-1)x-(m+3)y-(m-11)=0, that is, the fixed point (2,3).

Let m = -3 get x = 2 and then let m = get y = 3, so the straight line passes through the fixed point (2, 3) and substitutes x = 2, y = 3 to get the left = 4m - 2-3m - 9-m + 11 = 0 = right, and it is proved.

1. The answer should be -10

Because the set of solutions to this inequality is (-1 2, 1 3). So the two solutions to the equation ax +bx+2=0 are -1 2 and 1 3

Then we can use the root-coefficient relationship x1+x2=-b a x1*x2=c a (c is 2.).With this equation, we can directly solve a=-12, and then substitute it into 1 to find b=-2)

2. It should be 1 2

Because x,y r,2x+y=2. According to the fundamental inequality. It can be found that 2xy is less than or equal to 2. under 2 times the root numberSo 2xy under the root number is less than or equal to 1

So xy is less than or equal to 1 2So c max is 1 2

3. It should be.

Because f(x) is an odd function. And because he is a subtractive function in (0, so he is monotonically increasing in (- 0).

Because it is necessary to make x·f(x) 0 true.

So x and f(x) must be different. Answer:

1.-1 2 and 1 3 are both solutions to the equation and take them in and solve a=-12 b=-2 so a-b=-10 c

2.The basic inequality is 2x+y=2 greater than or equal to 2 * the root number is 2xy, so xy is less than or equal to 1 2 b

3 is obtained by the odd function f(3)=0 when x(3 + infinity) (infinity. -3) xf(x) less than 0 c

Solution: ab=5, bc=7

bc：ab=7：5

i.e., the ratio is 1 and 2 fifths

The tolerance of the two equal difference series is 4,6 respectively, the tolerance of the new series is the least common multiple of 4,6 12, and the first common term is 2, so the new series is the first term is 2, the tolerance is 12 equal difference series, the last term 182 = 12 15 + 2 is the 16th term, so the sum of the slip terms of the new series is (2 + 182) 16 2 = .

Look at the Nian Feng model boy slow Tu Ji late.

From the sine theorem a sina = c sinc, so c = 60 or 120 degrees.

When c = 60, a = 30, so b = 90, from the sine theorem b = 6

When c=120, a=30, so b=30, isosceles triangle, b=a=3

Sine theorem: a sina = c sinc, i.e. 3 1 2 = 3 times the root number 3 sinc

So sinc=root number 3 2, so c=60° or 120° When c=60°, b is equal to 90°, at this point b=6

When c = 120°, b is equal to 30 degrees, and b = 3

Combination of sine theorem and cosine theorem, basic questions.

Use the sinusoidal theorem to find c = 60 degrees or 120 degrees, 60: that is, the right triangle b = 90 degrees, b = 6

120: isosceles triangle b=a=3

1）q^4-q^2=72

q^2（q^2-1）=72

q^2=9q=±3

2) When q=3, all items in the series are positive, and -81 is not an item in the series;

q=-3, an=(-3) n, -81 is also not an term in the series, and in summary, -81 is not an term in the series.

Solution: First of all, all three edges must be greater than zero, and we get: a 0

a+1＞0a+2＞0

Solution: a 0

Secondly, the acute angle triangle should ensure that the three angles are less than 90°, that is, the largest angle should be less than 90°, and the large angle can be matched according to the large side, and the angle of A+2 is the largest.

That is, the cosine value of the angle to which a+2 is greater than zero, let this maximum angle be a

According to the cosine theorem:

cosa=[a +(a+1) -a+2) ] 2a(a+1) 0 simplification (a-3)(a+1) 2a(a+1) 0 because a 0

So A-3 0

Solution a 3

Because of the triangle.

So a+a+1>a+2

Get a>1

Because of acute angles.

The angle of a+2 is the largest, and the angle is less than 90 degrees, let the angle be acosa=(a2+(a+1)2-(a+2)2) (2a(a+1))=(a-3) (2a)>0

Get a>3 or a<0

In summary, A>3

a+a+1>a+2 gets a>1, a+a+2>a+1 gets a>-1: that is, a>1 is an acute triangle, so a 2+(a+1) 2-(a+2) 2>0, a 2-2a-3>0 is a>3 or a<-1 and finally intersects, a>3