"Equations" Equality and Inequality PPT (Properties of Equations and Solutions to Equations in Lesson 1)

Description

"Equations" Equality and Inequality PPT (Properties of Equations and Solutions to Equations in Lesson 1)

Part One: Learning Objectives

Master the properties of equations and be able to use the cross multiplication method to decompose factors

Be able to use the properties of equations to solve linear equations of one variable, and be able to use factoring methods to solve quadratic equations of one variable

Equation PPT, part 2 content: independent learning

Problem guide

Preview the contents of textbook P43-P46 ​​and think about the following questions:

1. What are the properties of equations?

2. What is the concept of identity?

3． What is the content of cross multiplication?

4. What is the concept of solution set of an equation?

A preliminary exploration of new knowledge

1. Properties of equations

(1) Add (subtract) ________ numbers or algebraic expressions to both sides of the equation at the same time, and the equation still holds;

(2) If both sides of the equation are multiplied (divided) by the same ________ number or algebraic expression at the same time, the equation still holds.

[Attention]The conditions for the establishment of the properties of equality, especially the "not zero" in property (2).

2. Identity

Generally speaking, for an equation containing letters, if the equation holds true when the letters in it take __________, it is called an identity, and it is also called _________ on both sides of the equation.

3． solution set of equation

Generally, the set composed of ________ of an equation is called the solution set of this equation.

self-test

Judge whether it is true or false (mark “√” if it is correct and “×” if it is wrong)

(1) If a=b, then a-c=b-c.()

(2) If a=b, then ac=bc.()

(3) If ac=bc, then a=b.()

(4)x3＋1＝(x＋1)(x2－x+1)． ()

(5)x2+5x+6=(x+2)(x+3)． ()

The deformation of the following expressions from left to right is ()

A． a2-b2+1=(a+b)(a-b)+1

B． m2-4m+4＝(m-2)2

C. (x+3)(x-3)＝x2-9

D. t2+3t-16=(t+4)(t-4)+3t

Equation PPT, the third part of the content: interactive teaching and practice

Decompose univariate polynomials using the cross multiplication method

Angle 1: Factorization of x2＋(p+q)x＋pq type

Factoring:

(1)x2-3x+2;

(2)x2＋4x－12.

[Solution] (1) As shown in the figure, decompose the quadratic term x2 into the product of the two x's in the figure, and then decompose the constant term 2 into the product of -1 and -2, and the two The sum of the products of the numbers is -3x, which is the linear term in x2-3x+2, so x2-3x+2=(x-1)(x-2).

regular method

The characteristics of quadratic trinomials such as x2＋(p+q)x＋pq are:

(1) The coefficient of the quadratic term is 1;

(2) The constant term is the product of two numbers;

(3) The linear term coefficient is the sum of two factors of the constant term.

Its decomposition factor is: x2+(p+q)x+pq=(x+p)(x+q).

Angle 2: Factorization of ax2＋bx＋c type

Factoring:

(1)6x2+5x+1;

(2)6x2+11x-7;

(3)42x2-33x+6;

(4)2x4－5x2+3.

regular method

For ax2+bx+c, decompose the coefficient a of the quadratic term into a1×a2, decompose the constant term c into c1×c2, and arrange a1, a2, c1, c2 as shown in the figure:, cross and multiply by diagonal lines, and then add, We get a1c2+a2c1. If it is exactly equal to the linear term coefficient b of ax2+bx+c, then ax2+bx+c can be decomposed into (a1x+c1)(a2x+c2), where a1 and c1 are located in the upper row in the above figure, and a2 and c2 are located in the next row.

Factoring bivariate polynomials using the cross multiplication method

Angle 1: Factorization of x2＋(p+q)xy＋pqy2

Factor the following expressions:

(1)a2-2ab-8b2;

(2)x+5xy-6y(x>0, y>0);

(3)(x＋y)2－z(x＋y)－6z2;

(4)m4＋m2n2－6n4.

regular method

The characteristics of quadratic homogeneous expressions such as x2＋(p＋q)xy＋pqy2 are:

(1) The coefficient of x2 is 1;

(2) The coefficient of y2 is the product of two numbers (pq);

(3) The coefficient of xy is the sum of these two numbers (p+q).

x2+(p+q)xy+pqy2=x2+pxy+qxy+pqy2=x(x+py)+qy(x+py)=(x+py)(x+qy).

Angle 2: Factorization of ax2＋bxy＋cy2 type

Factor the following expressions:

(1)6m2-5mn-6n2;

(2)20x2+7xy-6y2;

(3)2x4+x2y2-3y4;

(4)6(x+y)+7z(x+y)+2z(x>0, y>0, z>0).

Equation PPT, Part 4: Feedback on Compliance

1. Factor x3-x, the result is ()

A． x(x2－1)B． x(x－1)2

C. x(x+1)2 D． x(x+1)(x－1)

2. It is known that a+b=3, ab=2, calculation: a2b+ab2 is equal to ()

A． 5B. 6

C. 9 D. 1

3． Factor a2+8ab-33b2 to get ()

A． (a+11)(a-3) B． (a+11b)(a-3b)

C. (a-11b)(a-3b) D. (a-11b)(a+3b)

4. The solution set of equation 3x(x-2)=2-x is ________.

5. Factor the following expressions:

(1)x2+15x+56;

(2)6x2+7x-3;

(3)x2-6xy-7y2;

(4)8x2+26xy+15y2.

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Update Time: 2024-06-28

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