Multiple choice
1. The following statement is true ().
A. The sum of a monomial and a monomial is a monomial.
B. the sum of a single term and a single term is a polynomial.
C. the sum of polynomials is polynomials.
D. the sum of algebraic expression and algebraic expression is algebraic expression.
Answer: d
Analysis: x2 and x3 are both monomials. The sum of two monomials x3 and x2 is x3+x2 is a polynomial, except a. Two monomials x2, the sum of 2x2 is 3x2 is a monomial, and B is excluded. The sum of the two polynomials x3+x2 and x3-x2 is 2x3, which is a monomial, excluding C, so D is chosen.
2. If both A and B represent rational numbers and a+b=0, then ()
A.a and b are both 0.
One of B.a and b is 0.
C.a and b are opposites.
D.a and b are reciprocal.
Answer: c
Analysis: Let a=2, b=-2, and satisfy 2+(-2)=0, so A and B are reciprocal.
3. The following statement is incorrect ()
A. there is the smallest natural number
B. There is no minimum positive rational number
C. negative integer without
D. non-negative numbers without
Answer: c
Analysis: The negative integer of is-1, so C is wrong.
4. If A and B represent rational numbers, and the value of a+b is greater than that of a-b, then ().
A.a and b have the same number.
B.a and b are different numbers.
C.a & gt0
D.b & gt0
Answer: d
5. An integer greater than-π and not a natural number is ()
A.2
B.3
C4
D. countless
Answer: c
Analysis: On the number axis, it is easy to see that the integers to the right of-π and to the left of 0 (including 0) are only -3, -2,
-1,0 * * * 4. Choose C.
6. There are four kinds of statements:
A. the square of a positive number is not necessarily greater than itself;
The cube of b positive number is not necessarily greater than itself;
C. the square of a negative number is not necessarily greater than itself;
The cube of a negative number is not necessarily greater than itself.
Among these four statements, the incorrect statement is ()
A: 0
B. 1
C.2
D.3
Answer: b
Analysis: the square of a negative number is a positive number, so it must be greater than itself, so C is wrong.
7.a stands for rational number, so the relationship between A and -a is ().
A.a is greater than-a.
B.a is less than-a.
C.a is greater than -a or a is less than-a.
D.a is not necessarily greater than-a.
Answer: d
Analysis: let a=0, you can immediately exclude a, b, c, and choose D.
8. In the process of solving the equation, in order to make the obtained equation and the original equation have the same solution, you can add () on both sides of the original equation.
A. multiply by the same number
B. Multiply by the same algebraic expression
C. Add the same algebraic expression
D. add both 1.
Answer: d
Analysis: For the deformation of the same solution of the equation, both sides of the equation are required to be multiplied by a number not equal to 0, so A is excluded. When we consider the equation x-2=0, it is easy to know that its root is x=2. If both sides of the equation are multiplied by an algebraic expression x- 1, we get (x- 1)(x-2)=0, its root is x= 1, and x=2. If it is not the same solution of the original equation, B is excluded. In fact, a constant is added to both sides of the equation, and the new equation has the same solution as the original equation. For d, the constant added here is 1, so choose d.
9. There was more than half a glass of water in the cup, which decreased by 10% on the second day and increased by 10% on the third day. So, the comparison result between the water in the cup on the third day and the first day is ().
A. also
Too much
C.it's gone
D.it may be more or less.
Answer: c
Analysis: Let the original amount of water in the cup be a, which can be obtained according to the meaning of the question.
The next day, the amount of water in the cup is a× (1-10%) = 0.9a;
On the third day, the amount of water in the cup is (0.9a) × (1+10%) = 0.9 ×1.1× a;
The ratio of that wat in the third cup to the water in the first cup is 0.99:1,
So on the third day, there was less water in the cup than on the first day, so C was chosen.
10. This ship goes back and forth between two docks of a river. If the speed of the ship itself in still water is fixed, then when the flow speed of the river increases, the time it takes for the ship to make a round trip will be ().
A. Improve
B. Reduce
C. constant D. possible increase or decrease
A: A.
fill-in-the-blank question
1. 1989 19902- 1989 19892=______。
Answer:198919902-198919892.
=( 1989 1990+ 1989 1989)×( 1989 1990- 1989 1989)
=( 1989 1990+ 1989 1989)× 1=39783979。
Analysis: Calculate with formula a2-b2=(a+b)(a-b).
2. 1-2+3-4+5-6+7-8+…+4999-5000=______。
Answer:1-2+3-4+5-6+7-8+…+4999-5000.
=( 1-2)+(3-4)+(5-6)+(7-8)+…+(4999-5000)
=-2500。
Analysis: This question uses the associative law in the operation.
3. When a=-0.2 and b=0.04, the value of algebraic expression a2-b is _ _ _ _ _.
Answer: 0
Analysis: The original formula ==(-0.2)2-0.04=0. Just substitute the known conditions into algebraic calculation.
4. There is 60 kilograms of salt water with 30% salt, which is evaporated on the scale. When salt water becomes 40% salt, the weight of salt water is _ _ _ _ _ grams.
Answer: 45,000 (grams)
Analysis: 60 kg of salt water with 30% salt content is 60×30% (kg).
Assuming that the weight of water evaporated to 40% salt is x grams,
That is, 0.00 1x kg, at this time 60×30%=(0.00 1x)×40%.
Solution: x=45000 (grams).
When we encounter this kind of problem, we need to find an invariant. The salt content in this problem is an invariant, and we can calculate it by listing the equations.
answer the question
1. Both parties have the same annual income. Party A saves 0.5 of 65438+ annual income every year, and Party B spends 100 yuan more than Party A every month. After three years, 600 yuan was in debt. What is everyone's annual income?
Answer: Assume that each person earns X yuan per year, and A starts from 4/5X yuan per year. According to the meaning of the question, there are:
3(4/5X+ 1200)=3X=600
(3- 12/5)X=3600-600
Solution, x=5000
A: Each person earns 5000 yuan a year.
2. If s =15+195+19995+19995+199.5 (44 9s), what is the sum of the last four digits of the sum number S!
Answer: S = (20-5)+(200-5)+(2000-5)++(2000-5) (45 zeros)
= 20+200+2000+200 0 (45 zeros) -5 * 45
= 22 20 (45 2)-225
= 22 2 1995 (42 2)
3. A person goes uphill at a speed of 3 km/h and downhill at a speed of 6 km/h, and the journey 12 km * * * It takes 3 hours and 20 minutes to try to find the distance between uphill and downhill.
Answer: Let's assume that the uphill journey is X kilometers and the downhill journey is Y kilometers.
x+y = 12①； x/3+y/6=3 1/3②
From ② 2x+y=20, ③
If ① has y= 12-x, substitute ③ to get 2x+ 12-x=20.
So x=8 (km), so y=4 (km).
Answer: 8 kilometers uphill and 4 kilometers downhill.
4. Prove that the remainder obtained by dividing the prime number p by 30 must not be a composite number.
Proof: let p=30q+r, 0 ≤ r.