Middle and high school mathematics knowledge bridging questions, middle and high school math bridgin

1. Let the two roots be x1 and x2, then.

x1+x2=m-1

x1x2=m

x1/x2=2/3

m=6,x1=2,x3=3

or m=1 6, x1=-1 3, x2=-1 2, so m=6 or m=1 6

2. Let f(x)=2x 2-6x+5=2(x-3 2) 2+1 2 maximum = f(0)=5

Minimum = f(3 2) = 1 2

Let f(x)=1 3x 2+2x=(x+3) 2 3-3 max=f(-5)=-5 3

Minimum value = f(-4) = -8 3

3. Proof: y=x 2-(m 2+4)x-2m 2-12x 2-(m 2+4)x-2(m 2+6)=(x-2)(x+m 2+6).

Let y=0, i.e., (x-2)(x+m 2+6)=0, x1=2, x2=-m 2-6

So the parabola must have a heel (2,0).

Let x1-x2=2+m2+6=12, i.e., m=2 or m=-2x1-x2=2+m2+6=m2+8, so when m=0, the minimum distance between the two fulcrums is 8

4. That is, the two roots of the equation kx2-2x+6k=0 are x=-3, x=-2x1+x2=-5=2 k

k=-2/5

The solution of the inequality is that x is not equal to 1 k, i.e., k<0, =4-4k*6k=0k=- 6 6

Use Vedic theorem and quadratic inequality and you're fine.

The first question is an old question, and the practice is also very difficult to think, this is the proof given by the old textbook Factorization:

a^3+b^3+c^3-3abc

a+b+c）(a^2+b^2+c^2-ab-ac-bc)=1/2（a+b+c）[(a-b)^2+(b-c)^2+(c-a)^2]

Prove that because a, b, and c are all positive numbers, so

1/2（a+b+c）[(a-b)^2+(b-c)^2+(c-a)^2] >= 0

Question 2 x 2 + 1 = 3x

Divide x 4+1+2x 2=9x 2 by dividing x 2 by x 2 to get x 2+1 x 2=7 and divide x by x to get x+1 x=3 x 3+1 x 3

x+1/x)(x^2+1/x^2)-(x+1/x)=21-3=18

Articulation point? Well, to tell you what is used in junior high school mathematics to high school mathematics, that is, knowledge extension, junior high school theoretical knowledge "real numbers" related, radical related concepts (open square simplification and deformation), power operation, the concept of power, absolute value operation, these are the need to understand and consolidate, the basic requirement is to try to be familiar with the operation proficiently.

Unary quadratic equations, the key of the key !!

1) The most practical method of finding the root: factorization (also called cross multiplication), formula method (you must know that it is derived by matching "perfect square", and you will know "why it can't be less than 0" when you understand it).

2) Regarding cross multiplication and Veda's theorem, if you don't master this, don't learn the images and analytic formulas of the quadratic function in high school math (this is more complicated for students, just look at the picture table to be familiar).

Nothing else, it's not interesting.

x 4-13x 2+9, first divided into two quadratic terms, and then continue to divide.

4x^4-4x^2)-(9x^2-9)=4x^2(x^2-1)-9(x^2-1)=(x^2-1)(4x^2-9)

x+1)(x-1)(2x+3)(2x-3)2, consider using the pending coefficient method.

Since 3x +5xy-2y = (3x-y)(x+2y), let 3x +5xy-2y +x+9y-4=(3x-y+a)(x+2y+b).

3x +5xy-2y +ax+2ay+3bx-by+ab=3x +5xy-2y +(a+3b)x+(2a-b)y+ab comparison coefficient, a+3b=1, 2a-b=9, ab=-4 solution a=4, b=-1

So 3x +5xy-2y +x+9y-4=(3x-y+4)(x+2y-1).

3. It cannot be decomposed within the range of rational numbers.

The breakdown of the range of real numbers is as follows:

x²-5x+3

x²-5x+25/4-25/4+3

x-5/2)²-13/4

x-5/2+√3/2)(x-5/2-√3/2)

Use " " for "divide".

1) The first method: divide into three cases, remove the absolute value, and solve.

The second method: imagine that there is a point x on the number line, and the distance of x from point 2 is subtracted from the distance of x from point -4. The x to the right of 2 gets the result of a maximum of 6, the left of -4 gets the result of a minimum of -6, and the x between -4 and 2 gets a result of a positive or negative 0, think for yourself...

This method is the combination of numbers and shapes, the combination of algebra and geometry, and the college entrance examination will be examined.

2) Using the above method of combining numbers and shapes, imagine that there is a point x on the number line, the distance of x from point 1 plus the distance of x from point -2. The minimum value is the point where x is between -2 and 1, there is no maximum value.

3) Multiply both sides of the equation by 2 to get: 2a + 2b + 2c = 2ab + 2bc + 2ca

Shift: a -2ab + b + b -2bc + c + a -2ca + c = 0

That is: (a-b) +b-c) +a-c) =0

I think that the squares are all greater than or equal to 0, so there must be a=b=c

4) x -y = 2xy both sides divided by y and move the term, (x y) 2 * x y -1=0, i.e. [ x y) -1 ] 2=0

Get x y = plus or minus root number 2

x-y) (x+y) The numerator denominator is divided by a number y (the equation does not change) at the same time, and (x y -1) (x y +1) is obtained.

5)(3a-5ab+3b) (5a+3ab+5b) The numerator denominator is divided by ab at the same time.

3/a + 3/b - 5) / (5/a + 5/b +3) =(3*2 - 5) / (5*2 +3) =1/13

6) The formula is of this form: 1 [a * a+2 )]= 1 2 * a+2)-a] [a * a+2 )]=1 2* [1 a -1 (a+2)].

Then 1x1/3 = 1 1 -1 3 divided by 2

2x1/4 = 1 2 -1 4 divided by 2

3x1/5 = 1 3 - 1 5 divided by 2

9x1/11 = 1 9 - 1 11 divided by 2

So 1x1/3 + 2x1/4 + 3x1/5 + ......1/9x11

The first and second questions are of one type, 1Mark two points on the number line, 2 and 4, the formula can be understood as the distance from any point to 2 on the number line minus the distance to 4, draw the number line and you can see that when x is on the right side of 2, the value of the formula is a maximum of 6, and when x is on the left side of 4, the value of the formula is a minimum of 6

2.It's the same as question 1, except that when x is between 2 and 1, the formula can be taken to a minimum value of 3, but the formula can't take the maximum value, you can draw a number line to know.

3.You can multiply the two sides of the known condition by 2 to give this equation: (a b) +b c) +c a) =0

So, a b, b c c a

4.From x -y = 2xy, we get: (x y) 2y So, x y 2·y x 1 2 y

Substitute the desired formula to get the answer.

5.From 1 a 1 b 1 2, you get 2ab a b, and you can substitute this into the original form to get 1 13.

The sixth question uses the sum of split terms commonly used in high school.

(1) You can make a piecewise function, when x is less than -4, the function y=6, when x is less than or equal to 2 and greater than or equal to -4, the function y=-2x-2, and when x is greater than 2, the function y=-6

By making an image, it can be observed: Maximum: 6 Minimum: -6 (2) There are various methods; The first method can be used, and the other is direct judgment, which can be derived according to the theorem 丨a+b丨 丨a丨 丨b丨.

3) Multiply the two sides by 2 and move the right side to the left: (a-b) 2+(a-c) 2+(b-c) 2=0, so a=b=c

4) Simplify the equation to x=(root number 2+1)y or x=(-root number 2+1)y, and substitute it to obtain it.

5) When a=b=1, the equation holds.

1. When x is greater than or equal to 2, the original formula is x-2-x-4=-6

When x is less than 2 and greater than or equal to 4, the original formula is x 2 x 4 2x 2 At this time, the original formula is always less than 6 and greater than 2

When x is less than 4, the original is x 2 x 4 6

So Big: 6Small: -6

2. When x is greater than or equal to 1, the original formula is y x 1 x 2 3x 1 and there is no maximum value.

When x is less than 1 and greater than or equal to 2, the original formula is y x 1 x 2 3

When x is less than 2, the original formula is y x 1 x 2 2x 1 Evergrande 3

So the minimum value is 3

3. A + b + c = ab + bc + ca is a ab + b bc + c ca 0

i.e. a(a b) b(b c) c(c a) 0 due to abc greater than 0

a b c so is an equilateral triangle.

Fourth, because x -y = 2xy, (x y) 2y 0

i.e. (x y root number 2 y) (x y root number 2 y) 0

So x root number 2 y y or x y root number 2 y

Substituting the value of x into x-y of x+y gives the root number two-1 or the negative root number two-1

5. 1 a 1 b 2 is divided into (a b) ab 2 i.e. a b 2ab

3a-5ab+3b）／（5a+3ab+5b）＝【3（a＋b）-5ab】／【5（a＋b）+3ab】＝1／13

Sixth, I didn't expect a simple algorithm.

Set -40

Primitive = 2-x-x-4=-2x-2 When x = -4 There is a maximum value of 6 to set x<-4

Original = 2-x + x + 4 = 6

Let x>2

Original = x-2-x-4=-6

So the maximum value is 6, and the minimum value is -6

y=4x^2 -4ax+a^2-2a+2

2x-a)*(2x-a)-2a+2---Eq. 1) From Eq. 1, we can see that the image of y is a parabola with an opening upward symmetry axis of a2;

So long to discuss in three situations.

The first case: when the axis of symmetry a 2>=2;

In this case, when x=2 is obtained, the minimum value is obtained, and x=2 is brought into the equation of y, 4x 2 -4ax+a 2-2a+2

That is, 4*2 -4ax+a 2-2a+2=3;

Solve a=5 + root number 10 or a=5 - root number 10;

Since a 2 > = 2, a = 5 + root number 10;

In the second case, when the axis of symmetry 0 is now x=a2, the minimum value is obtained, and x=a2 is brought into the equation of y 4x2 -4ax+a2-2a+2

i.e. -a+2=3;

Solve a=5;

Due to 0 the third case: when the axis of symmetry a 2<0;

In this case, when x=0 is obtained, the minimum value is obtained, and x=0 is brought into the equation of y, 4x 2 -4ax+a 2-2a+2

i.e. a -2a + 2 = 3;

Solve a=1 + root number 2 or 1 - root number 2;

Since a 2<0, a = 1 - root number 2;

If the symmetry axis of the function is x=a 2, then it is discussed in different cases:

1.a 2 0; Finally, we get a=1-2 or a=5+10

The axis of symmetry is x=a 2, the quadratic function of the opening upward, when a 0, the minimum value of f(x) is f(0)=a -2a+2=3, and the solution obtains a=-1 + root number 2 [round] or a=-1 - root number 2, when 0 a 4, the minimum value of f(x) is f(a 2) = -2a + 2 = 3, and the minimum value of f(x) when a 4 is f(2)=16-8a+a -2a+2=3 to obtain a=5-root number 10 [round] or a=5 + root number 10. In summary, we get a = -1 - root number 2 or a = 5 + root number 10