Find the formulas you need to use in middle school mathematics Theorems

The math formula for the second year of junior high school is as follows.

Multiplication and factorization, a2-b2=(a+b)(a-b),a3+b3=(a+b)(a2-ab+b2),a3-b3=(a-b(a2+ab+b2).

Trigonometric inequality, |a+b|≤|a|+|b|，|a-b|≤|a|+|b|，|a|≤b-b≤a≤b，|a-b|≥|a|-|b|-|a|≤a≤|a|。

Solution of a quadratic equation, -b+ (b2-4ac) 2a, -b- (b2-4ac) 2a. The relationship between the root and the coefficient, x1+x2=-b a, x1*x2=c a.

Note: Vedic theorem, discriminant formula b2-4ac=0 Note: The equation has two equal real roots, b2-4ac>0 Note: The equation has two unequal real roots, b2-4ac<0 Note: The equation has no real roots, there are conjugate complex roots.

Binomial theorem.

a+b)^2=a^2+2ab+b^2 (1, 2, 1)(a+b)^3=a^3 + 3a^2*b + 3a*b^2 + b^3 (1, 3, 3, 1)

a-b)^3=a^3 - 3a^2*b + 3a*b^2 - b^3

a+b) 4= 1, 4, 6, 4, 1 (Baskar trigonometric coefficient) squared difference formula.

a-b)*(a+b)=a2 - b2 cube and cubic variance formula.

a-b)(a^2+ab+b^2)=a^3-b^3(a+b)(a^2-ab+b^2)=a^3+b^3Induction formulaRepresentation of angles in radians:

sin（2kπ+αsinα （k∈z）

cos（2kπ+αcosα （k∈z）

tan（2kπ+αtanα （k∈z）

cot（2kπ+αcotα （k∈z）

sec（2kπ+αsecα （k∈z）

csc（2kπ+αcscα （k∈z）

1. Doubling formula.

tan2a=2tana/（1-tan2a） ctg2a=（ctg2a-1）/2ctga

cos2a=cos2a-sin2a=2cos2a-1=1-2sin2a

2. Quadratic equation formula.

The equation is: ax2 bx c 0, b2 4ac is called the discriminant of the root, when there are two roots greater than 0, it is equal to 0 with two equal real roots, and less than 0, the equation has no real roots.

3. Function formula:

The formula for a primary function y kx b, the image of which is a straight line;

The inverse proportional function formula y k x, its image is hyperbola.

4. Quadratic function formula.

y=ax²+bx+c;(a, b, c are constants, a≠0), and its image is a parabola. y is called the quadratic function of x, the three elements of a parabola: the opening direction, the axis of symmetry, and the vertex.

5. The length of the inner male tangent = d-(r-r) and the outer male tangent length = d-(r+r).

6. Two circles are separated from d r+r

The two circles are circumscribed d=r+r

Two circles intersect r-r d r+r(r r).

The two circles are inscribed d=r-r(r r).

The two circles contain d r-r(r r).

1 There is only one straight line after two points 2 The line segment between two points is the shortest 3 The complementary angles of the same angle or equal angles are equal 4 The co-angles of the same angle or equal angles are equal5 There is only one straight line perpendicular to the known straight line after one point.

6 Of all the line segments where the outer point of the line is connected to the points on the line, the perpendicular line segment is the shortest7 The axiom of parallelism After passing through the outer point of the line, there is only one line parallel to the line8 If both lines are parallel to the third line, the two lines are also parallel to each other.9 The isotope angle is equal, the two lines are parallel 10 The inner angle is equal, the two lines are parallel11 The same side inner angle is complementary, the two lines are parallel 12 The two lines are parallel, the isotopic angle is equal13 The two lines are parallel, and the inner angle is equal 14 The two lines are parallel and the inner angles of the same side are complementary15 Theorem The sum of the two sides of a triangle is greater than the third side.

16 Corollary The difference between the two sides of a triangle is less than the third side.

17 The sum of the inner angles of a triangle Theorem The sum of the three internal angles of a triangle is equal to 180°18 Corollary 1 The two acute angles of a right triangle are mutually congruent.

19 Corollary 2 One outer angle of a triangle is equal to the sum of two interior angles not adjacent to it20 Corollary 3 One outer angle of a triangle is greater than any one of its inner angles that are not adjacent to it.21 The corresponding sides and corresponding angles of a congruent triangle are equal.

22 The axiom of corner edges There are two triangles with equal sides and their angles are congruent23 The axiom of corner corners There are two triangles with two angles and their angles corresponding to the equal congruence of two triangles24 Corollary There are two angles and the opposite side of one of the corners corresponds to the congruence of two triangles25 The axiom of edges has three sides corresponding to two triangles that are equal26 The axiom of hypotenuse and right angle side There are hypotenuse and one right side corresponding to two equal right triangle congruence.

27 Theorem1 The distance from a point on the bisector of an angle to both sides of the angle is equal28 Theorem2 To a point where both sides of an angle are at the same distance, on the bisector of this angle29 The bisector of an angle is the set of all points that are equally distant from both sides of the angle30 The property theorem of an isosceles triangle The two base angles of an isosceles triangle are equal31 Corollary1 The bisector of the apex of an isosceles triangle bisects the base edge and is perpendicular to the base.

In a right triangle, the opposite side of an angle of 30° is half the hypotenuse.

1) Use the formula method:

a2-b2=(a+b)(a-b)

a2+2ab+b2=(a+b)2

a2-2ab+b2=(a-b)2

If you reverse the multiplication formula, you can use it to factor certain polynomials. This method of factoring is called using formulas.

b) Square Difference Formula.

1 square difference formula.

1) Equation: a2-b2=(a+b)(a-b)(3) factorization.

1. When factoring, if there is a common factor for each item, the common factor should be mentioned first, and then further decomposed.

2 Factorization, which must be carried out until the factors of each polynomial can no longer be factored.

d) Perfect square formula.

1) Reverse the multiplication formula (a+b)2=a2+2ab+b2 and (a-b)2=a2-2ab+b2 to get :

a2+2ab+b2 =(a+b)2

a2-2ab+b2 =(a-b)2

There are many more, and you have to learn to tidy up yourself.

So many, it is estimated that it is dark when listed.

That's what your question looks like......This is OK with the perfect square formula......9（a-b)²+12(a²+b²）+4（a+b）²=（3（a-b））²3×2×2（ a+b）（a-b）+（2（a+b））²

3（a-b）+2（a+b））²

3a-3b+2a+2b)²

5a-b)²

You can think of a-b and a+b as x, y

Original = 9x 2 + 12xy + 4y 2 = (3x+2y) 2

Substitution=(5a-b) 2