7 hằng đẳng thức đáng nhớ và hệ quả cùng các dạng toán

Bạn đang xem: 7 hằng đẳng thức đáng nhớ và hệ quả cùng các dạng toán
tại TRƯỜNG THCS TT PHÚ XUYÊN

7 memorable equality constants and consequences and math forms

7 memorable equality constants and their consequences and types of math students have learned in the Math 8 program, Algebra. This knowledge is quite important in the program, related to many other types of equation solving. To know more about the knowledge to remember, please share the following article!

I. THEORY OF THE 7 MEMORABLE AWAKENING LEADES

1. What are the seven memorable equality constants?

You’re viewing: 7 memorable equality constants and consequences and math forms

The seven memorable equality constants are the most basic equations that every math student needs to master. The equality is proved by multiplying polynomial by polynomial. These rows of equality are in the group of elementary algebraic equality rows, among many others.

These equations are frequently used in problems related to solving equations, multiplying and dividing polynomials, and transforming expressions at the middle and high school levels. Memorizing seven memorable equality constants helps to quickly solve problems of factoring polynomials.

2. Seven memorable equality constants and consequences

a. Seven memorable equality constants:

Square of a sum:

The square of a sum is equal to the square of the first number plus twice the product of the first number times the second, plus the square of the second
Square of a difference:

The square of the difference is equal to the square of the first number minus twice the product of the first number times the second number then plus the square of the second number.
Difference of two squares:

The difference between the squares of two numbers is equal to the sum of the two numbers multiplied by the difference of the two numbers.
The cube of a sum:

The cube of a sum of two numbers is equal to the cube of the first number plus three times the product of the square of the first number times the second plus three times the product of the first number times the square of the second plus the cube of the second .
The cube of a difference:

The cube of a difference of two numbers is equal to the cube of the first number minus three times the product of the square of the first number times the second plus three times the product of the first number times the square of the second minus the cube of the number. Monday
Sum of two cubes:

The sum of two cubes of two numbers is equal to the sum of the two numbers multiplied by the missing square of the difference
Difference of two cubes:

The difference of two cubes of two numbers is equal to the difference of the two numbers multiplied by the missing square of the sum of the two numbers.

b. Consequences of equality

In addition, we have the consequent equality constants of the above seven equality constants. Often used while transforming trigonometry to prove equality, inequality, etc.

Consequences with the 2nd degree equality constant

$(a+b)^2=(a-b)^2+4ab$

$(a-b)^2=(a+b)^2-4ab$

$a^2+b^2=(a+b)^2-2ab$

$(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc$

$(a+b-c)^2=a^2+b^2+c^2+2ab-2ac-2bc$

$(a-b-c)^2=a^2+b^2+c^2-2ab-2ac-2bc$

Consequences with the 3 degree equality constant

$a^3+b^3=(a+b)^3 - 3ab(a+b)$

$a^3-b^3=(ab)^3 + 3ab(ab)$

$(a+b+c)^3=a^3+b^3+c^3+3(a+b)(a+c)(b+c)$

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

$(ab)^3+(bc)^3+(ca)^3=3(ab)(bc)(ca)$

$(a+b)(b+c)(c+a)-8abc=a(bc)^2+b(ca)^2+c(ab)^2$

$(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc$

$(a+b)(b+c)(c+a)-8abc=a(bc)^2+b(ca)^2+c(ab)^2$

$(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc$

General consequences

$a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}b^2-a^{n-4}b ^3+ldots+a^2b^{n-3}-acdot b^{n-2}+b^{n-1})$

$a^nb^n=(ab)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+ldots+a^2b^{n-3}+ ab^{n-2}+b^{n-1})$

3. Some notes on memorable equality constants

+ Transforming equality constants is mainly the way to transform from sum, product difference between numbers, skills of polynomial analysis to factorization must be mastered in order to apply the new equality constants clearly and accurately.

+ To better understand the nature of using equality, when applied to problems, students can prove that the existence of equality constants is true by reverse transformation, using the related isomorphisms into proving the problem.

+ While using equality constants in algebraic fractions, students should keep in mind that there will be many forms of distortion of the formula due to the nature of each problem, but the essence is still the above formulas, just the transform back and forth to fit in the calculation.

For example :

4. Quick tips for memorizing equality constants

If we pay attention, we can see that the equality constants 1 and 2, 4 and 5, 6 and 7 are all quite similar, only differing in sign, so the thing to note here is to write remember their marks, so that we can memorize them correctly and without confusion.

For equality constants 5 and 6, it should be noted that:

“The sum of the cubes is equal to the product of the sum of two numbers and the missing square of the difference”

“The difference of cubes is equal to the product of the difference of two numbers and the missing square of a sum”

In addition, you can collect more songs about “Seven Memorable Constants” created by author “Nhat Anh” based on the music of the song “After All”. Surely, when listening to this song, students can both relax and remember knowledge in the most natural way without feeling dry and confusing.

However, all of this only contributes to a small part, the most important thing is that you need to understand its nature and constantly practice, practice and practice hard, you will easily remember it.

II. FREQUENTLY FREQUENT MATH FORMS

Form 1: Find the minimum value of the expression For example: Calculate the minimum value of the expression: A = x2 – 2x + 5* The answer:– We have: A = x2 – 2x + 5 = (x2 – 2x + 1) + 4 = (x – 1)2 + 4– Because (x – 1)2 ≥ 0 for all x.⇒ (x – 1) 2 + 4 ≥ 4 or A ≥ 4– So the smallest value of A = 4, The sign “=” occurs when: x – 1 = 0 or x = 1⇒ Conclusion Amin = 4 ⇔ x = 1Form 2: Find the maximum value of the expressionFor example: Calculate the maximum value of the expression: A = 4x – x2* The answer:– We have: A = 4x – x2 = 4 – 4 + 4x – x2 = 4 – (4 – 4x + x2) = 4 – (x2 – 4x + 4) = 4 – (x – 2)2– Because (x) – 2)2 ≥ 0 for all x ⇔ -(x – 2)2 0 for all x⇔ 4 – (x – 2)2 4 [cộng 2 vế với 4]⇔ A ≤ 4 The sign “=” occurs when: x – 2 = 0 or x = 2⇒ The conclusion that A’s net worth is: Amax = 4 ⇔ x = 2.

Form 3: Calculate the value of the expressionFor example: Calculate the value of the expression: A = x2 – 4x + 4 at x = -1* The answer.– We have: A = x2 – 4x + 4 = x2 – 2.x.2 + 22 = (x – 2)2– At x = -1 : A = ((-1) – 2)2=(-3 )2= 9⇒ Conclusion: So at x = -1, then A = 9 Form 4: Prove that expression A does not depend on the variable For example: Prove that the following expression does not depend on x: A = (x – 1)2 + (x + 1)(3 – x)* The answer.– We have: A =(x – 1)2 + (x + 1)(3 – x) = x2 – 2x + 1 – x2 + 3x + 3 – x = 4 : constant does not depend on variable x.

Form 5: Proof of equality For example: Prove that the following equality is true: (a + b)3 – (a – b)3 = 2b(3a2 + b2)* The answer:– For this math we transform VT = VP or VT = A and VP = A– We have: VT = (a + b)3 – (a – b)3= (a3 + 3a2b + 3ab2 + b3) – (a3 – 3a2b + 3ab2 – b3)= a3 + 3a2b + 3ab2 + b3 – a3 + 3a2b – 3ab2 + b3= 6a2b + 2b3= 2b(3a2 + b2) = VP (dpcm).⇒ Conclusion, then : (a + b)3 – (a – b)3 = 2b(3a2 + b2)• Form 6 : Proving the inequality– Transform the inequality into the form of the expression A ≥ 0 or A ≤ 0. Then use the transformations to bring A to 1 of the 7 equality constants.For example: Prove that expression B takes a negative value for all values of the variable x, knowing: B = (2-x)(x-4)-2* The answer: – We have: B = (2-x)(x-4) – 1 = 2x – 8 – x2 + 4x – 2 = -x2 + 6x – 9 – 1 = -(x2 – 6x + 9) – 1 = – (x-3)2 – 1– Since (x-3)2 ≥ 0 ⇔ -(x-3)2 ≤ 0 ⇒ -(x-3)2 – 1 ≤ -1 < 0 for all x,

Form 7: Find the value of xFor example:Find the known value of x: x2( x – 3) – 4x + 12 = 0* The answer.x2 (x – 3) – 4x + 12 = 0⇔ x2 (x – 3) – 4(x – 3) = 0⇔ (x – 3) (x2 – 4) = 0⇔ (x – 3)(x –) 2)(x + 2) = 0⇔ (x – 3) = 0 or (x – 2) = 0 or (x + 2) = 0⇔ x = 3 or x = 2 or x = –2⇒ Conclusion, so solution : x = 3; x = 2; x = -2

Form 7: Factoring polynomials

For example:Factor the following polynomial: A = x2 – 4x + 4 – y2* Solution:– We have: A = x2 – 4x + 4 – y2 [để ý x2 – 4x + 4 có dạng hằng đẳng thức]= (x2 – 4x + 4) – y2 [nhóm hạng tử]= (x – 2)2 – y2 [xuất hiện đẳng thức số A2 – B2]= (x – 2 – y )( x – 2 + y)⇒ A = (x – 2 – y )( x – 2 + y)

III. EXERCISES ABOUT THE CONDITIONAL STANDARDS

Lesson 1: Quick calculation

$1. 1001^{2}$

2. 29.9.30.1

$3. 201^{2}$

4. 37.43

$5. 199^{2}$

$6. 37^{2}+2.37 .13+13^{2} \$

$7. 51.7-2.51,7.31,7+31.7^{2} \$

$8. 20,1.19.9 \$

$9. 31.8^{2}-2.31,8.21.8+21.8^{2} \$

$10.33.3^{2}-2.33,3.3,3+3,3^{2}\$

Lesson 2: Simplify and then calculate the value of the expression

$1. (mathrm{x}-10)^{2}-mathrm{x}(mathrm{x}+80)$

$2. (2 mathrm{x}+9)^{2}-mathrm{x}(4 mathrm{x}+31)$

$3. 4 mathrm{x}^{2}-28 mathrm{x}+49$

$4. mathrm{x}^{3}-9 mathrm{x}^{2}+27 mathrm{x}-27$

Lesson 3: Prove for every integer N the expression $(2 n+3)^{2}-9$ divisible by 4

Exercise 4: Write the following expression in the form of product

$a. (x+y+x)^{2}-2(x+y+x)(y+z)+(y+z)^{2}$

$b. (x+y+x)^{2}-(y+z)^{2}$

$c. (x+3)^{2}+4(x+3)+4$

$d. 25+10(x+1)+(x+1)^{2}$

Lesson 5. Write the following expression in the form of product

$a. x^{2}-2$

$b. y^{2}-13$

$c. 2 x^{2}-4$

$d. left(x^{2}-1right)^{2}-(y+3)^{2}$

$e. left(a^{2}-b^{2}right)^{2}-left(a^{2}+b^{2}right)^{2}$

Lesson 6. Write the following expression in the form of product

$a. -4 x^{2}+9 y^{2}$

$b .8+(4 x-3)^{3}$

Lesson 7. Write the following expression as sum

$a. (x+y+z+t) cdot(x+y-z-t)$

b.. $(x+2 y+3 z+t)^{3}.$

Exercise 8: Write the following expression as a sum

$a. left(x^{2}-2 x-1right)^{2}$

b. $left(m^{2}+2 m-3right)^{2}.$

$text { c. }(x+1)left(x^{2}+1right)left(x^{4}+1right)$

$d.2. (3+1)left(3^{2}+1right)left(3^{4}+1right)$

****Advanced equations problems (with answers)

Lesson 1. Given the polynomial 2x² – 5x + 3 . Write the above polynomial as a polynomial of the variable y where y = x + 1.

The answer

According to the problem we have: y = x + 1 => x = y – 1.

A = 2x² – 5x + 3

= 2(y – 1)² – 5(y – 1) + 3 = 2(y² – 2y + 1) – 5y + 5 + 3 = 2y² – 9y + 10

Lesson 2. Quickly calculate the results of the following expressions:

a) 127² + 146.127 + 73²

b) 98.28– (184 – 1)(184 + 1)

c) 100² – 99² + 98² – 97² + …+ 2² – 1²

d) (20² + 18² + 16² +…+ 4² + 2²) – ( 19² + 17² + 15² +…+ 3² + 1²)

The answer

a) A = 127² + 146.127 + 73²

= 127² + 2.73.127 + 73²

= (127 + 73)²

= 200²

= 40000 .

b) B = 9 8 .2 8 – (18 4 – 1)(18 4 + 1)

= 188 – (188 – 1)

= 1

c) C = 100² – 99² + 98² – 97² + …+ 2² – 1²

= (100 + 99)(100 – 99) + (98 + 97)(98 – 97) +…+ (2 + 1)(2 – 1)

= 100 + 99 + 98 + 97 +… + 2 + 1

= 5050.

d) D = (20² + 18² + 16² +…+ 4² + 2²) – ( 19² + 17² + 15² +…+ 3² + 1²)

= (20² – 19²) + (18² – 17²) + (16² – 15²)+ …+ (4² – 3²) + (2² – 1²)

= (20 + 19)(20 – 19) + (18 + 17)(18 – 17) + (16 + 15)(16 – 15)+ …+ (4 + 3)(4 – 3) + (2 + 1)(2 – 1)

= 20 + 19 + 18 + 17 + 16 +15 + …+ 4 + 3 + 2 + 1

= 210

Lesson 3. Compare the following two numbers, which is greater?

a) A = (2 + 1)(22+1)(24+1)(28 + 1)(216 + 1) and B = 232

b) A = 1989.1991 and B = 19902