quadratic radical

Knowledge point 1. The key of quadratic root: master the concept of quadratic root. Difficulties: the meaningful conditions of quadratic roots

formula

(a≥0) is called quadratic radical.

Knowledge point 2. The simplest quadratic root

Key point: master the conditions of the simplest secondary root [source: learning. Difficulties: correctly distinguish whether it is the simplest quadratic root.

At the same time, it meets the following requirements: ① the factor of the root sign is an integer and the factor is an algebraic expression (the denominator does not contain the root sign); (2) The number of roots contains factors or factors that can be completely opened. Such a quadratic root is called the simplest quadratic root.

Knowledge point 3. Similar quadratic root

Key points: mastering the conceptual difficulties of similar quadratic roots: correctly distinguishing whether they are similar quadratic roots.

After several quadratic roots are transformed into the simplest quadratic roots, if the number of roots is the same, these quadratic roots are called similar quadratic roots.

Knowledge point 4. Properties of quadratic radical

Key points: Mastering the nature of quadratic root Difficulties: Understanding and skillfully using the nature of quadratic root.

①(

)2 = a(a≥0)；

②

=│a│=

;

Knowledge point 5. Denominators are physical chemistry and physical chemistry factors.

Key points: master the concepts of denominator and factor.

Difficulties: Understand denominator skillfully and find rational factors.

Removing the root sign in the denominator is called denominator rationalization; Two algebras with quadratic roots are multiplied, and if their product does not contain quadratic roots, they are said to be mutually rational factors.

Example Observe the reasonable calculation of the following denominator:

, find out the law from the calculation results, and use this law to calculate:

=_____________

Think about solving problems:

Knowledge point 6. Square root operation

Key points: Mastering the algorithm difficulty of secondary roots: Skillfully operating secondary roots.

External shift and internal shift of (1) factors: if some factors in the root sign can be fully expanded, they can be replaced by their arithmetic roots and moved outside the root sign; If the root sign is in the form of algebraic sum, first solve the factor and convert it into the form of product, then move the factor out of the root sign, and vice versa, square the positive factor outside the root sign and move it into the root sign.

(2) Addition and subtraction of quadratic roots: first, transform the quadratic roots into the simplest quadratic roots, and then merge similar quadratic roots.

(3) Quadratic root multiplication and division method: Quadratic root multiplication (division) and square root multiplication (division), the product (quotient) obtained is still the square root of the product (quotient) and the operation result is turned into the simplest quadratic root.

=

(a≥0，b≥0)；

(b≥0，a & gt0).

(4) additive commutative law, associative law, multiplicative commutative law and associative law of rational numbers, distributive law of multiplication to addition and polynomial multiplication formula are all applicable to the operation of quadratic roots.

Requirements and proposition trends in the latest exam question 1 Master the relevant knowledge of quadratic roots, including concepts, properties, operations, etc. 2. Skillfully perform the operation of quadratic root.

One-dimensional quadratic equation

I. Knowledge structure:

Quadratic equation of one variable: concept, solution and solution, practical application, relationship between roots and coefficients.

Second, a detailed analysis of the test sites

Test center 1. Concept definition (1): ① contains only one unknown, ② the highest degree of the unknown is 2, so ③ the integral equation is a quadratic equation with one variable.

(2) General expression:

⑶ Difficulties: How to understand "the highest number of unknowns is 2": ① The coefficient is not "0"; ② The unknown index is "2";

(3) If there is an index to be determined or a coefficient to be determined, it is necessary to establish an equation or inequality to discuss.

Example 2, Equation

Is a quadratic equation about x, then the value of m is.

Test site 2, the solution of the equation

⑴ Concept: The value of the unknown quantity that makes both sides of the equation equal is the solution of the equation. ⑵ Application: Using the concept of root to find the value of algebraic expression;

Typical example: for example 1, it is known.

The value of is 2, then

The value of is.

Test point 3, solution

⑴ Methods: ① Direct opening method; ② Factorization method; ③ Matching method; ④ Formula method (1) Emphasis: order reduction.

1, direct open method:

For. ※.

,

The direct opening method is applicable to all other forms.

Typical example: example 1, solving equation:

=0;

Example 2, if

The value of x is.

2, factorization type:

Characteristics of the equation: the left side can be decomposed into the product of two linear factors, and the right side is "0". ※,

Equation form: For example ※

,

,

Typical example: for example 1,

The root of is () a.

B.

c。

D.

Example 2, if

The value of 4x+y is.

Type 3, matching method

When solving equations, collocation method is often not used. But the idea of formulas is often used to solve problems such as the value or extreme value of algebra. ※.

Typical example: interpretation by trial matching method

The value of is always greater than 0.

4. Formula method (1) Conditional expression:

(2) Formula:

,

Typical example: for example 1. Choose an appropriate method to solve the following equation:

⑴

⑵

⑶

Type 5, the application of "descending thinking"

(1) Find the value of algebraic expression; ⑵ Solve binary quadratic equation.

Typical example: known

, find algebraic expression.

The value.

Test site 4. Discrimination of roots

The function of the discriminant of roots: ① to determine the number of roots; ② Find the value of undetermined coefficient; ③ Suitable for others.

Typical example: example 1, if about

Equation of

If there are two unequal real roots, the value range of k is.

Test site 5. "Classification Discussion" in Equation Problems

Typical example: for example 1. Discuss the equation about x.

The situation of the root.

Test site six, the application of problem solving

(1) the problem of "meeting"; (2) the problem of "compound interest"; (3) the problem of "geometry";

(4) the problem of "maximum"; 5) "Chart" problem

Typical example:

1. Cut a 20 cm long iron wire into two sections and make a square around the length of each section.

(1) What are the lengths of these two wires to make the sum of the areas of these two squares equal to 17cm2?

Test site seven. Relationship between root and coefficient

(1) Premise: For

Generally speaking, when ① is satisfied.

、②

When,

With Vieta's theorem.

② Main contents:

⑶ Application: Overall substitution evaluation.

Typical example: for example 1. The equation about x is known.

There are two unequal real roots.

,

(1) Find the value range of k;

(2) Is there a real number k that makes the two real numbers of the equation in opposite directions? If it exists, find the value of k; If it does not exist, please explain why.

Radial

Knowledge network diagram

pattern plan

Identification and application

Coordinates of points symmetrical about the origin.

Central symmetry

Centrally symmetric figure

Graphic rotation

Translation and attributes

Translation and attributes

Rotation and attributes

( 1)

Central symmetry: rotate the figure around a certain point.

If it can coincide with another figure, this point is called the center of symmetry, and the corresponding points in these two figures are symmetrical about this point.

(2)

On the nature of rotation: the distance from the corresponding point to the center of rotation is equal; The included angle between the corresponding point and the connecting line of the rotation center is equal to the rotation angle; Graphic consistency before and after rotation.

Question 1. The following is a centrally symmetric graph ()

(1) line segment; (2) Angle; (3) equilateral triangle; (4) square; (5) parallelogram; (6) rectangle; (7) isosceles trapezoid.

A.2 B.3 C.4 D.5

Answer: C.

Question 5. In a line segment, a ray, two intersecting straight lines and a five-pointed star, the number of centrally symmetric figures is ().

A. 1 B.2 C.3 D.4 answer: B.

circle

First, knowledge points

1, angles related to the circle-central angle and circumferential angle.

The central angle ∠ AOB in (1) graph; Angle of circle

ACB；

(2) As shown in the figure, if ∠AOB=50 degrees and ∠ACB= 25 degrees.

Degree;

(3) In the above figure, if AB is the diameter of circle O, then ∠AOB= 180.

Degree; ∠ACB= 90

Degree;

2, the symmetry of the circle:

The (1) circle is an axisymmetric figure, and its symmetry axis is arbitrary.

A straight line passing through the center of the circle;

A circle is a figure with a symmetrical center, and the symmetrical center is the center of the circle.

(2) Vertical diameter theorem: the diameter perpendicular to the chord bisects the chord and bisects the arc opposite the chord.

As shown in the figure, ∫CD is the diameter of circle O, and CD⊥AB is in E∴ =, =

3. There are three positional relationships between a point and a circle: a point is in a circle, a point is in a circle, and a point is in a circle;

4. There are three positional relationships between a straight line and a circle: phase, phase and phase.

5, the position relationship between the circle and the circle:

6, tangent properties:

Example 4: (1) As shown in the figure, if PA is the tangent of ⊙O and point A is the tangent point, then ∠PAO= degrees.

(2) As shown in the figure, PA and PB are tangents of ⊙O, and points A and B are tangents.

Then =, ∞ =∞;

7. Calculation in the circle.

Calculation formula of (1) arc length:

Example 5: If the central angle of a sector is 60 and the radius is 3, what is the arc length of this sector?

Solution: Because the arc length of the sector =

therefore

=

= (The answer is still π)

(2) Sector area:

Example 6: ① If the central angle of a sector is 60 and the radius is 3, what is the area of this sector?

Solution: Because the area of the sector is S=

So S=

= (The answer is still π)

② If the arc length of the sector is 12πcm and the radius is 6㎝, what is the area of this sector?

Solution: Because the area of the sector is S=

So S= =

(3) Cone:

Example 7: If the generatrix length of a cone is 5cm and the radius is 4cm, what is the lateral area of the cone?

Solution: The side development diagram of a cone is a shape, and the arc length of the development diagram is equal to.

Transverse area of cone =

Probabilistic preliminary

Knowledge carding

1. Random events in life are divided into definite events and uncertain events, and definite events are divided into inevitable events and impossible events. Among them,

① The probability of inevitable events is 1, that is, p (inevitable events) =1;

② The probability of an impossible event is 0, that is, p (impossible event) = 0;

③ If A is an uncertain event, then 0

2. Calculation method of random event probability:

(1) theoretical calculation is divided into the following two cases:

The first type: the probability of random events involving only one-step experiment, such as: calculating a kind of probability model according to the relationship between the size and area of probability;

The second type: calculate the probability of random events involving two or more experiments by list method, enumeration method and tree diagram, such as calculating whether the game is fair or not.

② The experimental estimation is divided into the following two situations:

The first method: probability estimation through experiments. We should know that when the number of experiments is very large, the experimental frequency can be used as an estimate of the probability of events, that is, a large number of experimental frequencies are stable to the theoretical probability.

The second method: probability estimation through simulation experiment. For example, using a calculator to generate random numbers to simulate experiments.

To sum up, the available probability models can be roughly divided into three categories; The first kind of problem has no theoretical probability, and its estimated value can only be obtained through experimental simulation; Although the second kind of problem has theoretical probability, it is not available at present, and its estimated value can only be obtained by experimental simulation. The third kind of problem is simple classical probability, and it is easy to calculate its probability in theory.