The first volume of seventh grade mathematics from formula to equation test questions
I. Multiple choice questions (* *11small questions)
1. Given m= 1, n=0, the value of algebraic expression m+n is ().
A.﹣ 1 b 1 c.﹣2 d . 2
2. Given x2 ~ 2x ~ 8 = 0, the value of 3x2 ~ 6x ~ 18 is ().
C.﹣ 10 D.﹣ 18
3. Transform the equation into x=2 according to ().
A. Properties of Equation 1 B. Properties of Equation 2
C. Basic properties of fractions D. Properties of inequalities 1
4. Given x2-2x-3 = 0, the value of 2x2-4x is ().
A.-6 b.6 c.-2 or 6 d.-2 or 30
5. If m-n =- 1, the value of (m-n) 2-2m+2n is ().
D.﹣ 1
6. Given x = 3, the value of 4-x2+x is ().
65438 AD+0 BC
7. According to the operation program shown in the figure, the output result can be 3, and the values of x and y are ().
a.x=5,y=﹣2 b.x=3,y=﹣3 c.x=﹣4,y=2 d.x=﹣3,y=﹣9
8. If m+n =- 1, the value of (m+n) 2-2m-2n is ().
1 D.2
9. Given that x-2y = 3, the value of algebraic expression 6-2x+4y is ().
B.﹣ 1 C.﹣3 D.3
10. When x= 1, the value of algebraic expression AX3-3bx+4 is 7, and when x =- 1, the value of this algebraic expression is ().
D.﹣7
1 1. The figure is a schematic diagram of the operation procedure. If the initial input value of x is 8 1, the output result of the 20th14th time is ().
A.3 B.27 C.9 D. 1
Fill in the blanks (* *18)
12. Given that the solution of equation 3a-x =+3 about x is 2, the value of algebraic expression a2-2a+ 1 is.
13. it is known that x=2 is the solution of equation a(x+ 1)= a+x about x, then the value of a is.
14. Follow the operation steps as shown in the figure. If the input value is 3, the output value is.
15. If a-2b = 3, then 2a-4b-5 =.
16. If m2-m = 6, then1-2m2+2m =.
17. When x= 1, the algebraic expression x2+ 1=.
18. If m+n=0, then 2m+2n+ 1=.
19. Calculate according to the program shown in the figure. If the input value of x is 3, the output value is.
20. According to the operation steps as shown in the figure, if the input value of x is 2, the output value is.
2 1. Given that the solution of equation 2x+A-5 = 0 about x is x=2, then the value of a is.
22. Liu Qian's magic show is popular all over the country. Xiaoming also learned that Liu Qian invented a magic box. When any real number pair (a, b) enters it, it will get a new real number: a2+b﹣ 1. For example, if (3, ﹣2) is put in, 32+ (.
23. If the value of x= 1 times number expression 2ax3+3bx+4 is 5, then the value of algebraic expression 2ax3+3bx+4 is.
24. If x2 ~ 2x = 3, the values of algebraic expressions 2x2 ~ 4x+3 are.
25. If m2 ~ 2m ~1= 0, the value of algebraic expression 2m ~ 4m+3 is.
26. Given x(x+3)= 1, the value of algebraic expression 2x2+6x-5 is.
27. Given x2 ~ 2x = 5, the value of algebraic expression 2x2 ~ 4x ~ 1 is.
28. The following is a simple numerical operation program. When the input value of x is 3, the output value is. (calculate or write with a scientific calculator).
29. There is a digitizer whose principle is shown in the figure. If the value of x is 7 at the beginning, we can find that the output result of 1 is 12, the output result of the second time is 6, and the output result of the third time is to continue in turn? The output result of 20 13 is.
Third, answer questions (* *1small questions)
30. Known: a=, b = |-2 |,. Find the value of algebraic expression: a2+b | 4c.
The seventh grade mathematics first volume test questions from formula to equation reference answer
I. Multiple choice questions (* *11small questions)
1. Given m= 1, n=0, the value of algebraic expression m+n is ().
A.﹣ 1 b 1 c.﹣2 d . 2
Algebraic evaluation of test points.
Analysis can substitute the values of m and n into algebraic formula to get the solution.
Solution: When m= 1 and n=0, m+n= 1+0= 1.
So choose B.
This topic reviews algebraic evaluation, and it is relatively simple to substitute the values of m and n.
2. Given x2 ~ 2x ~ 8 = 0, the value of 3x2 ~ 6x ~ 18 is ().
C.﹣ 10 D.﹣ 18
Algebraic evaluation of test points.
Special calculation problems.
After analyzing the first two terms of the equation and extracting three deformations, the known deformation of the equation can be calculated and the numerical value can be obtained.
Solution: ∵x2﹣2x﹣8=0, which means x2﹣2x=8.
3x2﹣6x﹣ 18=3(x2﹣2x)﹣ 18=24﹣ 18=6.
So choose B.
Comment on this question to examine algebraic evaluation, using the idea of whole substitution, is a basic question type.
3. Transform the equation into x=2 according to ().
A. Properties of Equation 1 B. Properties of Equation 2
C. Basic properties of fractions D. Properties of inequalities 1
Properties of test site equation.
Analysis According to the basic properties of the equation, the original formula can be analyzed.
Solution: according to the properties of equation 2, the equation is transformed into x = 2；;
Therefore, choose: B.
Comments on this topic mainly examine the basic properties of the equation. The property of the equation is: 1. When the same number or letter is added or subtracted from both sides of the equation, the equation is still valid. 2. When both sides of the equation are multiplied or divided by the same non-zero number or letter, the equation still holds.
4. Given x2-2x-3 = 0, the value of 2x2-4x is ().
A.-6 b.6 c.-2 or 6 d.-2 or 30
Algebraic evaluation of test points.
The overall idea of the topic.
Multiply both sides of the analytical equation by 2 at the same time, and then calculate 2x2 ~ 4x.
Solution: x2-2x-3 = 0.
2? (x2﹣2x﹣3)=0
2? (x2﹣2x)﹣6=0
2x2﹣4x=6
Therefore, choose: B.
Comment on this topic. The key to solve the problem is to get the required 2x2 ~ 4x.
5. If m-n =- 1, the value of (m-n) 2-2m+2n is ().
D.﹣ 1
Algebraic evaluation of test points.
Special calculation problems.
After analyzing the formula, two items are extracted and deformed, and the value of m﹣n is substituted into the calculation to get the numerical value.
Solution: ∵ m ∵ n = ?1,
(m﹣n)2﹣2m+2n=(m﹣n)2﹣2(m﹣n)= 1+2=3.
So choose: a.
Comment on this question to examine algebraic evaluation, using the idea of whole substitution, is a basic question type.
6. Given x = 3, the value of 4-x2+x is ().
65438 AD+0 BC
Algebraic evaluation of test site: mixed operation of scores.
Special calculation problems.
After analyzing the last two terms of the formula, the common factor is deformed, the known equation is named and deformed, and then the numerical value is obtained by calculation.
Solution: ∫ x = 3,
x2﹣ 1=3x
x2﹣3x= 1，
The original formula = 4-(x2-3x) = 4-=.
Therefore, choose: d.
This topic reviews the evaluation of algebraic expressions, and it is the key to solve this problem to properly deform the expressions of known sum and solution.
7. According to the operation program shown in the figure, the output result can be 3, and the values of x and y are ().
a.x=5,y=﹣2 b.x=3,y=﹣3 c.x=﹣4,y=2 d.x=﹣3,y=﹣9
Algebraic evaluation of test site: solution of binary linear equation.
Special calculation problems.
The analysis lists the equations according to the operation program, and then according to the definition of the solution of the binary linear equation, the analysis and judgment of each option are solved by the exclusion method.
Solution: From the meaning of the question, 2x-y = 3,
When a, x=5, y=7, so option a is wrong;
B, x=3, y=3, so option b is wrong;
C, X =-4, Y =- 1 1, so the c option is wrong;
When d and X =-3 and Y =-9, the d option is correct.
Therefore, choose: d.
This topic reviews algebraic evaluation, mainly using the solution of binary linear equation, and understanding the operation program to list equations is the key to solving problems.
8. If m+n =- 1, the value of (m+n) 2-2m-2n is ().
1 D.2
Algebraic evaluation of test points.
The overall idea of the topic.
Analysis takes (m+n) as a whole and substitutes it into algebraic expression to get the solution.
Solution: ∫ m+n =1
(m+n)2﹣2m﹣2n
=(m+n)2﹣2(m+n)
=(﹣ 1)2﹣2? (﹣ 1)
= 1+2
=3.
So choose: a.
This topic reviews algebraic evaluation, and the application of the overall idea is the key to solving the problem.
9. Given that x-2y = 3, the value of algebraic expression 6-2x+4y is ().
B.﹣ 1 C.﹣3 D.3
Algebraic evaluation of test points.
First, transform 6 ~ 2x+4y into 6 ~ 2 (x ~ 2y), and then substitute x ~ 2y = 3 into the whole calculation.
Solution: ∫ x2y = 3,
6﹣2x+4y=6﹣2(x﹣2y)=6﹣2？ 3=6﹣6=0
So choose: a.
This topic comments on algebraic expression evaluation: first, the algebraic expression is deformed according to the known conditions, and then the whole idea is used for calculation.
10. When x= 1, the value of algebraic expression AX3-3bx+4 is 7, and when x =- 1, the value of this algebraic expression is ().
D.﹣7
Algebraic evaluation of test points.
The overall idea of the topic.
Substitute x= 1 into the algebraic expression to get the relationship between a and b, and then substitute X =- 1 into the calculation to get the solution.
Solution: when x= 1, ax3-3bx+4 = a-3b+4 = 7,
The solution is a-3b = 3,
When AX3 ﹣ 3bx+4 = ﹣ a+3b+4 = ﹣ 3+4 =1.x=﹣ 1
So choose: C.
This topic reviews algebraic evaluation, and the application of the overall idea is the key to solving the problem.
1 1. The figure is a schematic diagram of the operation procedure. If the initial input value of x is 8 1, the output result of the 20th14th time is ().
A.3 B.27 C.9 D. 1
Algebraic evaluation of test points.
Topic chart type.
Analyze and calculate according to the operation program, and then get the law. From the fourth operation, the result of even operation is 1, the result of odd operation is 3, and then the solution is enough.
Solution: 1,? 8 1=27,
The second time? 27=9,
The third time? 9=3,
The fourth time? 3= 1,
The fifth time, 1+2=3,
The sixth time? 3= 1,
,
By analogy, the result of even operation is 1, and the result of odd operation is 3.
20 14 is even,
The output result of 20 14 is 1.
Therefore, choose: d.
In commenting on this topic, we examine algebraic evaluation. According to the calculation program, the output result of even operation is 1, and the output result of odd operation is 3, which is the key to solve the problem.
Fill in the blanks (* *18)
12. Given that the solution of equation 3a-x =+3 about x is 2, the value of algebraic expression a2-2a+ 1 is 1.
Solutions of one-dimensional linear equations.
First, substitute x=2 into the equation to find the value of a, and then substitute the value of a into the algebraic expression for calculation.
Solution: ∵ The solution of equation 3a about x∵x =+3 is 2,
3a ~ 2 =+3, and a=2 is obtained.
The original formula is = 4-4+ 1 = 1.
So the answer is: 1.
This question examines the solution of a linear equation with one variable, and knowing the basic steps of solving a linear equation with one variable is the key to solving this question.
13. it is known that x=2 is the solution of equation a(x+ 1)= a+x about x, then the value of a is.
Solutions of one-dimensional linear equations.
Special calculation problems.
The value of a can be obtained by substituting x=2 into the equation.
Solution: Substitute x=2 into the equation to get 3a= a+2.
Solution: a=
So the answer is:
This topic reviews the solution of a linear equation with one variable. The solution of the equation is an unknown number that can make the left and right sides of the equation equal.
14. Follow the operation steps as shown in the figure. If the input value is 3, the output value is 55.
Algebraic evaluation of test points.
Topic chart type.
The analysis can be calculated according to the formula of the operation program to get the solution.
Solution: As can be seen from the figure, when the input value is 3, (32+2)? 5=(9+2)? 5=55.
So the answer is: 55.
This topic reviews algebraic evaluation, and understanding the topic operation program is the key to solving the problem.
15. If a-2b = 3, then 2a-4b-5 = 1.
Algebraic evaluation of test points.
Analysis transforms the algebraic expression into an algebraic expression in the form of (a-2b), and then substitutes a-2b = 3 as a whole and evaluates it.
Solution: 2A-4B-5
=2(a﹣2b)﹣5
=2? 3﹣5
= 1.
So the answer is: 1.
Comment on this topic and examine the evaluation of algebraic expressions. The numbers expressed by letters in algebraic expressions are not explicitly told, but are implied in the topic design. First, get the value of algebraic expression (a-2b) from the topic design, and then use the value of algebraic expression (a-2b). Integral replacement method Find the value of algebraic expression.
16.(20 13? Sunshine) If m2-m = 6, then1-2m2+2m =-11.
Algebraic evaluation of test points.
The overall idea of the topic.
In the analysis, m2-m is regarded as a whole and can be solved by substituting algebraic expression.
Solution: ∫ M2M = 6,
1﹣2m2+2m= 1﹣2(m2﹣m)= 1﹣2？ 6=﹣ 1 1.
So the answer is:-1 1.
This topic reviews algebraic evaluation, and the application of the overall idea is the key to solving the problem.
17. When x= 1, the algebraic expression x2+ 1= 2.
Algebraic evaluation of test points.
Analysis can substitute the value of x into algebraic expression to get the solution.
Solution: When x= 1, x2+1=12+1= 2.
So the answer is: 2.
This review examines algebraic evaluation, which is the basic problem and accurate calculation is the key to solving the problem.
18. If m+n=0, then 2m+2n+ 1= 1.
Algebraic evaluation of test points.
The algebraic expression is transformed into the form of known conditions, and then substituted into the whole calculation.
Solution: ∫m+n = 0,
2m+2n+ 1=2(m+n)+ 1，
=2? 0+ 1,
=0+ 1,
= 1.
So the answer is: 1.
This topic reviews algebraic evaluation, and the application of the overall idea is the key to solving the problem.
19. Calculate according to the program shown in the figure. If the input value of x is 3, the output value is -3.
Algebraic evaluation of test points.
Topic chart type.
According to the analysis, the value of x is odd, and the solution can be obtained by substituting the following relationship.
Solution: When x=3, the output value is -x =-3.
So the answer is: -3.
This topic reviews algebraic evaluation, and accurate selection of relational expressions is the key to solving problems.
20. According to the operation steps shown in the figure, if the input value of x is 2, the output value is 20.
Algebraic evaluation of test points.
Topic chart type.
Analyze the formula written according to the operation program, and then substitute the data for calculation to get the solution.
Solution: As can be seen from the figure, the operation program is (x+3) 2 ~ 5.
When x=2, (x+3) 2-5 = (2+3) 2-5 = 25-5 = 20.
So the answer is: 20.
This topic reviews algebraic evaluation, which is a basic topic. It is the key to solve the problem to write the operating program accurately according to the chart.
2 1. It is known that the solution of equation 2x+A-5 = 0 about x is x=2, so the value of a is 1.
Solutions of one-dimensional linear equations.
Substituting x=2 into the equation, we can get an equation about a, and just solve the equation.
Solution: Substitute x=2 into the equation and get: 4+a-5 = 0.
Solution: a= 1.
So the answer is: 1
This topic examines the definition of equation solution, and understanding the definition is the key.
22. Liu Qian's magic show is popular all over the country. Xiaoming also learned that Liu Qian invented a magic box. When any real number pair (a, b) enters it, it will get a new real number: a2+b﹣ 1. For example, if (3, ﹣2) is put in, 32+ (.
Algebraic evaluation of test points.
Special application problems.
Through analysis and observation, we can see that the value of unknown quantity is not directly given, but implicit in the problem, so we need to find out the law and substitute it for solution.
Solution: According to the given rule: m=(﹣ 1)2+3﹣ 1=3.
The final real number is 32+ 1- 1 = 9.
According to the rules, the value of m is calculated first, and then further calculation, which implies the overall mathematical thinking and correct operation ability.
23. If x= 1 and the value of algebraic expression 2ax3+3bx+4 is 5, then when x =- 1, the value of algebraic expression 2ax3+3bx+4 is 3.
Algebraic evaluation of test points.
X= 1 is substituted into the algebraic expression 2ax3+3bx+4, so that its value is 5 and the value of 2a+3b is obtained. Then substitute X =- 1 into the algebraic expression 2ax3+3bx+4, and calculate the deformed offspring to get the value.
Solution: When ∵x= 1, the algebraic expression 2ax3+3bx+4=2a+3b+4=5, namely 2a+3b= 1.
When x = |1,the algebraic expression 2ax3+3bx+4 = | 2a | 3b+4 = | (2a+3b)+4 = |1+4 = 3.
So the answer is: 3
Comment on this question to examine algebraic evaluation, using the idea of whole substitution, is a basic question type.
24. If x2 ~ 2x = 3, the algebraic expression 2x2 ~ 4x+3 has a value of 9.
Algebraic evaluation of test points.
Special calculation problems.
After analyzing the deformation of the first two terms of the formula, the known equation can be substituted into the calculation to get the numerical value.
Solution: ∫ x22x = 3,
2x2﹣4x+3=2(x2﹣2x)+3=6+3=9.
So the answer is: 9.
Comment on this question to examine algebraic evaluation, using the idea of whole substitution, is a basic question type.
25. If m2 ~ 2m ~1= 0, the value of algebraic expression 2m ~ 4m+3 is 5.
Algebraic evaluation of test points.
The overall idea of the topic.
In the analysis, the value of m2 ~ 2m is obtained first, then the algebraic expression is sorted into the form of known conditions, and the solution is obtained by substitution calculation.
Solution: m2-2m = 1 is obtained from m2-2m = 1.
So, 2m ~ 4m+3 = 2 (m2 ~ 2m)+3 = 2? 1+3=5.
So the answer is: 5.
This topic reviews algebraic evaluation, and the application of the overall idea is the key to solving the problem.
26. Given x(x+3)= 1, the algebraic expression 2x2+6x-5 has a value of -3.
Algebraic evaluation of test site: polynomial multiplied by monomial.
The overall idea of the topic.
The algebraic expression is sorted into the form of known conditions, and then the solution can be obtained by substituting the data for calculation.
Solution: ∫x(x+3)= 1,
2x2+6x﹣5=2x(x+3)﹣5=2？ 1﹣5=2﹣5=﹣3.
So the answer is: -3.
This topic reviews algebraic evaluation, and the application of the overall idea is the key to solving the problem.
27. Given x2 ~ 2x = 5, the algebraic expression 2x2 ~ 4x ~ 1 has a value of 9.
Algebraic evaluation of test points.
The overall idea of the topic.
Analysis arranges algebraic expressions into the form of known conditions, and then substitutes them for calculation to get the solution.
Solution: ∫ x22x = 5,
2x2﹣4x﹣ 1
=2(x2﹣2x)﹣ 1，
=2? 5﹣ 1,
= 10﹣ 1,
=9.
So the answer is: 9.
This topic reviews algebraic evaluation, and the application of the overall idea is the key to solving the problem.
28. The following is a simple numerical operation program. When the input value of x is 3, the output value is 1. (calculate or write with a scientific calculator).
Algebraic evaluation of test points.
Theme finale; Chart type.
When the input value of x is 3, the square is 9, plus (-2) is 7, and finally dividing by 7 equals 1.
Solution: Algebraic expression can be obtained from the title map: (x2-2)? 7.
When x=3, the original formula = (32-2)? 7=(9﹣2)? 7=7? 7= 1
So the answer is: 1
The comment on this question examines algebraic evaluation. This kind of problem should correctly express algebraic expressions, and then calculate algebraic values. The key to solve this problem is to find out the calculation program given by the problem.
29. There is a digitizer whose principle is shown in the figure. If the value of x is 7 at the beginning, we can find that the result of the 1 th output is 12, the result of the second output is 6, the result of the third output is 3, and so on? The output result of 20 13 is 3.
Algebraic evaluation of test points.
Theme finale; Chart type.
The analysis shows that the input X is 7, the output result is x+5, even number 12 is substituted into X to get the calculation result of 6, even number 6 is substituted into X to get the output result of the third time, and so on to get the general law, thus getting the result of the 20th time13.
Solution: According to the meaning of the question, the value of X is 7 at first, and the output result of 1 is 7+5 =12;
What is the result of the second output? 12=6;
What is the result of the third output? 6=3;
The result of the fourth output is 3+5 = 8;
What is the result of the fifth output? 8=4;
What is the result of the sixth output? 4=2;
What is the result of the seventh output? 2= 1;
The result of the eighth output is1+5 = 6;
The output result obtained by the induction and summary cycle starts from the second time with 6, 3, 8, 4, 2, 1.
∵(20 13﹣ 1)? 6=335? 2,
The output result of 20 13 is 3.
So the answer is: 3; three
This topic comments on algebraic evaluation, and understanding the laws in the topic is the key to solving this topic.
Third, answer questions (* *1small questions)
30. Known: a=, b = |-2 |,. Find the value of algebraic expression: a2+b | 4c.
Algebraic evaluation of test points.
Thematic calculation problems; The finale.
Analysis: Substitute the values of A, B and C into the calculation to get the numerical value.
Solution: when a=, b = |-2 | = 2, c=,
a2+b﹣4c=3+2﹣2=3.
Comment on this problem and examine algebraic evaluation, involving knowledge such as simplification of quadratic root, mixed operation of absolute value and rational number. Mastering the algorithm skillfully is the key to solve this problem.