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Give counter examples to prove that the following propositions are false propositions (1) if a > b, then 1 / A
(1) If a > b, then 1 / A0, that is, a and B have the same sign;
If x ^ 2-3x + 2 = 0, then x = 1 or x = 2
Be sure to remember the meaningful conditions
Please judge whether the following proposition is true or false. If it is false, please give counter examples. If it is true, please give proof
(1) if a > b, then a & # 178; > b & # 178;
(2) an isosceles triangle with an angle equal to 60 ° is an equilateral triangle
False-1 and 2
really
Simple calculation: 56 / (5.6 * 4) first; 8.4/9 + 1.6/4 second; 4.68 - (1.68-0.32) third; 7.3/4 + 2.7 * 0.25 fourth
If you want to write the formula right + reward 5, copy the second one wrong ~ 8.4/6 + 1.6/4
56/（5.6*4）=(56*10)/（5.6*4*10）=(56*10)/（56*4）=10/4=2.58.4/9+1.6/4=8.4/9+0.4/1=8.4/9+3.6/9=12/9=4/34.68-（1.68-0.32）=4.68-1.68+0.32=3+0.32=3.327.3/4+2.7*0.25=7.3/4+2.7*1/4=7.3/4+2.7/4=10/4=2.5...
For rational numbers a and B, the definition of operation is as follows: a ★ B = (a + b) (a-b), then the value of - 3 ★ 4
-3★4=（-3+4）（-3-4）=1×（-7）=-7．
The number of 5 digits which are composed of 1,2,3,4,5 and have no repetition and 1,2 are not adjacent to 5 is
Put all the five digits in the following order:
Remove the case that 1 is adjacent to 5: * 4!
Remove the case that 2 is adjacent to 5: * 4!
In addition, the repeated removal of 1 and 2 are adjacent to 5: * 3!
Get - 2! * 4! - 2! * 4! + 2! * 3! = 36
So the repeated group may be 1.52 or 2.51. Why only add 2! * 3! Don't multiply by 2?
The arrangement of 5 numbers: 5!
1. 5. Adjacent arrangement: 2!,
The adjacent (15) as a whole is arranged with 2, 3 and 4: 4!,
It's two times four!
2. 5. Adjacent arrangement: 2!,
The adjacent (25) as a whole is arranged with 1, 3 and 4: 4!,
It's two times four!
1. 2 and 5 are adjacent, 5 must be in the middle, there are: 2!,
The adjacent (125) as a whole is arranged with 3 and 4: 3!,
It's two times three!
The total number is: 5! - 2! × 4! - 2! × 4! + 2! × 3! = 36
Let a be an M * n matrix and prove that R (a) = m if and only if there exists an n * m matrix B such that ab = E
Adequacy:
Because R (a) = M
There are invertible matrices P of order m and Q of order n such that PAQ = [em, 0]
Let d = [em, 0] ^ t,
Then paqd = em, i.e. AQDP = em,
Let B = QDP, ab = em
The sufficiency is proved
necessity
It is known that there exists n * m matrix B such that ab = E
Let's suppose that for a, there exists an invertible matrix P of order m and an invertible matrix Q of order n such that PAQ = C=
【Er,0】
【0,0】
That is R (a) < M
A=P^(-1)CQ^(-1)
AB=P^(-1)CQ^(-1)B=E
CQ^(-1)BP^(-1)=E
Because the last M-R lines of C are all zero, R (a) = M
The necessity has been proved
If a two digit number is divided by one digit number, the quotient is still two digits and the remainder is eight. What is the sum of the divisor quotient and the remainder
We need it in a hurry
Divisor greater than remainder
The one digit number larger than 8 is only 9
So the divisor is 9
The smallest two digits are 10
If the quotient is 11
The divisor is 11 × 9 + 8 = 107, which is a three digit number
So the quotient is 10
Then the divisor is 10 × 9 + 8 = 98
So the sum of divisor quotient and remainder is 98 + 10 + 9 + 8 = 125
Primary school mathematics problem solving: 1111 = 02222 = 03333 = 04444 = 41234 = 15678 = 3, find 6666 =?
6666 = 4 1111 has no circle, so it's 0 2222 has no circle, and it's also 0 3333. Similarly, 4444 has four circles, so it's 4 1234 4, so it's 1 5678, 6 and 8, so there's one for each of the three 6666, so it's 4
This is the kindergarten promotion examination and civil service examination
Add brackets to the equation to the left of the following equal sign to make the equation true
-The quadratic power of 8 × 3 is - 6 = - 24
-Quadratic power of 4 + 12 △ 4 + 1 = 0
-1 / 2-1 / 3 + 1 / 4 × 12 = - 7
-The quadratic power of 4 - 4 × 3 = 4
-8 × (quadratic power of 3-6) = - 24
(- 4 quadratic + 12) △ 4 + 1 = 0
(- half - one third + one fourth) × 12 = - 7
The second power of (- 4) - 4 × 3 = 4
The approximate number is 0.1040______ The significant number is______ ．
The last digit of approximate number 0.1040 is 0 in ten thousand, accurate to ten thousand; the significant digits are 1, 0, 4, 0
RELATED INFORMATIONS
- 1. Judge whether the following propositions are true or false? If false, give counter examples; if true, give proof (1) An angle must be less than its complement (2) If two lines are parallel, the inner angles of the same side must not be equal
- 2. If AB = 0, then a = 0 is a false proposition
- 3. A counterexample is given to show that the following proposition is a false proposition, [1] if a > b, then 1 / A
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