The common factor method test questions in the second volume of mathematics in the first day of junior high school
First, multiple-choice questions (4 points for each small question, *** 12 points)
The common factor of 1. polynomial 8xmyn- 1- 12x3myn is ().
A.xmyn B.xmyn- 1
C.4xmyn D.4xmyn- 1
2. Observe the following types: ① ABX-ADX; ②2x2y+6xy 2； ③8 m3-4 m2+2m+ 1； ④a3+a2 b+ ab2-B3；
⑤(p+q)x2y-5x 2(p+q)+6(p+q)2； ⑥a2(x+y)(x-y)-4b(y+x)。 Among them, () can be factorized by increasing the common factor.
A.①②⑤ B.②④⑤
C.②④⑥ D.①②⑤⑥
3. (-8) 2014+(-8) 2013 divisible by the following number is ().
a3 b . 5 c . 7d . 9
Fill in the blanks (4 points for each small question, *** 12 points)
4.(20 13? Factorization: 32B-4AB =.
5.(20 13? Is (2x-2 1)(3x-7)-(3x-7) known? (x- 13) can be factorized into (3x+a)(x+b), where both a and b are integers, then a+3b=.
6. Calculation: (1)3.982-3.98? 3.97= .
(2)0.4 1? 25.5+0.35? 25.5+2.4? 2.55= .
Iii. Answering questions (***26 points)
7.(8 points) Try to explain that 8 17-279-9 13 will be divisible by 45.
8.(8 points) Factorize first, and then calculate and evaluate.
(1) (2x-1) 2 (3x+2)+(2x-1) (3x+2) 2-x (1-2x) (3x+2), where x =/kloc-2.
(2)5x(m-2)-4x (m-2), where x = 0.4 and m = 5.5.
Extension extension
9.( 10) Factorize (1), (2) and (3) first, and then answer the following questions.
( 1) 1+a+a( 1+a)。
(2) 1+a+a( 1+a)+a( 1+a)2。
(3) 1+a+a( 1+a)+a( 1+a)2+a( 1+a)3。
Question:
(1) Explore the law of factorization first, and then write1+a+a (1+a)+a (1+a) 2+a (1+a) 3+? The result of factorization of +a( 1+a)20 14.
② Please factorize according to the above method:1+a+a (1+a)+a (1+a) 2+a (1+a) 3+? +a( 1+a)n(n is a positive integer).
The answer to the math common factor test in the second volume of the first day of junior high school.
1.D. The common factor of polynomial 8xmyn- 1- 12x3myn is 4xmyn- 1.
2. Analytically select d.① abx-adx = ax (b-d); ②2x2y+6xy 2 = 2xy(x+3y)； ③8m3-4m2+2m+ 1 cannot be decomposed by common factor method; ④a3+a2b+ab2-b3 cannot be decomposed by common factor method;
⑤(p+q)x2y-5x 2(p+q)+6(p+q)2 =(p+q)[x2y-5x 2+6(p+q)]； ⑥a 2(x+y)(x-y)
4b(y+x)=(x+y)[a2(x-y)-4b]。 So 1256 can be decomposed by the common factor method.
3. The analysis choice is C. (-8) 2014+(-8) 2013.
=(-8)? (-8)20 13+(-8)20 13
=[(-8)+ 1] (-8)20 13
=(-7)? (-8)20 13
=820 13? 7.
So it can be divisible by 7.
4. Analytical formula =ab(3a-4).
Answer: ab(3a-4)
5. Analyze (2x-21) (3x-7)-(3x-7) (x-13) = (3x-7) (2x-21-x+13) = (.
Then a=-7, b=-8, a+3b=-7-24=-3 1.
Answer: -3 1
6. Analysis (1)3.982-3.98? 3.97
=3.98? 3.98-3.98? 3.97
=3.98? (3.98-3.97)=3.98? 0.0 1=0.0398.
(2)0.4 1? 25.5+0.35? 25.5+2.4? 2.55
=0.4 1? 25.5+0.35? 25.5+0.24? 25.5
=25.5? (0.4 1+0.35+0.2 4)
=25.5? 1=25.5.
Answer: (1)0.0398 (2)25.5
7. Analysis because 817-279-913 = (34) 7-(33) 9-(32)13.
=328-327-326=326(32-3- 1)
=326? 5=324? 32? 5=(32? 5)? 324=45? 324,
Is it because of 45 again? 324 is divisible by 45,
So 8 17-279-9 13 will be divisible by 45.
8. Analyze the original formula (1) = (2x-1) 2 (3x+2)+(2x-1) (3x+2) 2+x (2x-1) (3x+2).
=(2x- 1)(3x+2)(2x- 1+3x+2+x)
=(2x- 1)(3x+2)(6x+ 1)。
When x= 1, the original formula = (2-1) (3+2) (6+1) =1? 5? 7=35.
(2)5x(m-2)-4x(m-2)=(m-2)(5x-4x)= x(m-2)。
When x =0.4 and m = 5.5, the original formula =0.4? (5.5-2)=0.4? 3.5= 1.4.
9. Analytical formula (1) = (1+a) = (1+a) 2.
(2) The original formula = (1+a) [1+a (1+a)] = (1+a) =1
(3) original formula = (1+a) [1+a+a (1+a)+a (1+a) 2]
=( 1+a)( 1+a)[ 1+a+a( 1+a)]
=( 1+a)2( 1+a)( 1+a)
=( 1+a)4。
① According to the laws of (1), (2) and (3),1+a (1+a)+a (1+a) 2+a (1+a) 3+? +a( 1+a)20 14 =( 1+a)20 15。
② The original formula = (1+a) [1+a+a (1+a)+a (1+a) 2+? +a( 1+a)n- 1]
=( 1+a)( 1+a)[ 1+a+a( 1+a)+a( 1+a)2+？ +a( 1+a)n-2]
=( 1+a)2( 1+a)[ 1+a+a( 1+a)+a( 1+a)2+？ +a( 1+a)n-3]
?
=( 1+a)n- 1( 1+a)( 1+a)=( 1+a)n+ 1。
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