Middle school math multiplication formula part questions, junior high school multiplication formula

1. The minimum value of the fill-in-the-blank question is = 4;Reason: (x 0 5+4x+4)-4=(x+2) 2-4 and (x+2) 2 minimum is 0; 2、.(8) to the power of 2004 times (to the power of 2005 = 1 8;Rationale:

8）^2004×（1/8）^ 2005=8^ 2004×8（^-2005）=1/83、.Knowing ab 0 5=-3, then -ab(a 0 5b to the fifth power - ab 0 6-b) = 33 reason: -ab(a 0 5b to the fifth power - ab 0 6-b=-a 3b 6-a 2b 4-ab 2=(ab 2) 3-(ab 2) 2-ab 2, substitute ab 0 5=-3 to obtain.

4. If x gets 2n power = 2, then (2x to the 3n power) 0 5 = if 64 0 5x8 0 6 = 2 to the n power, then n = the first blank: 32, the second blank: 21 reason:

2x to the power of 3n) 0 5=4x 6n=4(x 2n) 3=4 2 3=32; 64 0 5x8 0 6=2 to the nth power = (2 6) 2 (2 3) 3=2 12 2 9=2 21 I don't have time now, let's do other questions next time.

4。.x�0�5+4x=.（x+2）�0�。

8 times the 2004 power (2005 times = (8 times 2004 times (. Substitute the part into parentheses first, and then substitute 32,211 after the calculation result. =(2x+y) 0 5(2x-y) 0 5(x+2y-1) 27 3m + 2n power = 3m x times x 2n power = x

We have learned some of the following multiplication formulas in junior high school:

1) Square Difference Formula.

Square Difference Formula.

2) Perfect square formula.

Perfectly squared formula.

Factorization. Factorization is an important identity deformation of algebraic formulas, and the common methods involved in junior high school textbooks mainly include: extracting common factors and formulas (square difference formula and perfect square formula), factorization and integer multiplication are deformations in the opposite direction, which play an important role in fractional operation, solving equations and various identity deformations, and is an important basic skill.

Multiplication formulas. <>

Formula. The integer deformation is an important algebraic identity deformation, and it is also an extremely common operation in high school mathematics

In junior high school, students are required to understand the concept of integers, and be able to do simple integer addition and subtraction operations, and multiplication operations (in which polynomial multiplication only refers to one-time multiplication); Simple calculations will be made using squared difference and perfect squared formulas; Factorization will be carried out by mentioning the common factor method and the Gong family method (directly using the formula no more than quadratic) for factorization (the exponent is a positive integer).

The monomial is multiplied by the polynomial: a(b+c)=ab+ac polynomial multiplied by polynomial: (a+b)(m+n)=am+am+bm+bn multiplied by the power of the base, the base does not change, and the exponents are added:

a m*a n=a (m+n) product of the power: (ab) n=a n*b n

Power to the power: (a n) m = a mn

Square difference formula: (a+b)(a-b)=a 2-b 2 perfect squared formula: (a+b) 2=a 2+2ab+b 2(a-b) 2=a 2-2ab+b 2

Cube sum formula: (a+b)(a2-ab+b 2)=a 3+b 3 cubic variance formula: (a-b)(a 2+ab+b 2)=a 3-b

Multiplicative distributive property: a(b+c)=ab+ac

Associativity: (a+b)(m+n)=am+am+bm+bn exponentiation: a m*a n=a (m+n).

The power of the product: (ab) n=a n*b n

Power to the power: (a n) m = a mn

Square difference formula: (a+b)(a-b)=a 2-b 2 perfect squared formula: (a+b) 2=a 2+2ab+b 2(a-b) 2=a 2-2ab+b 2

Cube sum formula: (a+b)(a2-ab+b 2)=a 3+b 3 cubic variance formula: (a-b)(a 2+ab+b 2)=a 3-b

Square difference formula: (a+b)(a-b)=a 2-b 2 perfect squared formula: (a+b) 2=a 2+2ab+b 2(a-b) 2=a 2-2ab+b 2

Cube sum formula: (a+b)(a2-ab+b 2)=a 3+b 3 cubic variance formula: (a-b)(a 2+ab+b 2)=a 3-b

Take a look at the equation (x+y) =x +2xy+y

Is it equal to the left and right?

2xy is because 2xy is not originally available, and it only has x +y in the problem, which is added by the solver to facilitate the solution. After adding 2xy, the equation is equal to the left and right. And then according to the conditions given, it is much better. Because x+y=a

xy=b, so (x+y) =a -2b. I don't know how to ask again.

Solution: (x 2-x-2)(bx-1) = bx 3-(b+1) x 2+(1-2b)x+2

x^3-2x^2+ax+1=bx^3-(b+1)x^2+(1-2b)x+2

b=1,a=-1

But I think there's something wrong with your question, and the remaining formula should be -1, otherwise it's not equal.

Pick B. n=3(x-3) 2-27+2(y+2) 2-8+35 =3(x-3) 2+2(y+2) 2>=0

Only true if x=3, y=-2, n=0. The rest of the cases are n>0

n=3x +2y -18x+8y+35=3x sedan chair-18x+27+2y closed wang +8y+8=3(x-3) 2+2(y+2) 2>=o

So Hu Que chooses B

B should be chosen. The original formula can be factored into 3 times (x-3) squared plus 2 times (y+2) squared, so it must not be negative.

Solution: Let the side of the original square be x cm.

x+3)^2-x^2=39

x^2+6x+9-x^2=39

x=5 If you are an elementary school student, draw a diagram and use arithmetic to solve it.

The original side length of this square is xcm

x+3)^2-x^2=39

x+3+x)(x+3-x)=39

3(2x+3)=39

2x+3=13

2x=10x=5

The original side length of the square is x

x+3)^2=x^2+39

x^2+6x+9=x^2+39

The original side length is x

by the title. x^2+39=(x+3)^2

Solution x=5

c.Square Difference Formula. a 2-b 2=(a+b)*(a-b) [2 is squared].

399 2=a+1 then a=399 2-1 according to the formula where a=399 b=1 so the answer is c