2. If the value of 2a+|4-5a|+| 1-3a| is constant, find the value range of a..
Because it is a fixed value, the unknown number A must have been dropped in the end, leaving only a constant term.
Observing the formula, the coefficients before a are 2, 3 and 5.
Considering whether the absolute value is positive or negative, there is no conclusion, but the 2a term must be positive-therefore, if you want to eliminate all a, you have to make 3a positive and 5a negative.
In other words:
Let |4-5a| open, and the absolute value will be negative, that is, 4-5a.
Solution: a
Similarly, | 1-3a| is positive after the absolute value is removed, that is,1-3a > 0
Solution: a> 1/3
At this time, what is the public formula A? The formula has a constant of 3.
At this time, the value range of a is: 1/3.
3, if 0
Classified discussion.
|x-5| >0, open absolute value, x=5+a, from 0.
|x-5|=0. Similarly, x=5 and a=0.
| x-5 | & lt； 0, similarly, 5-a=x,-10.
Integer value of a: from -4 to 9, and then from -9 to 4, a *** 19.
Add a: 0
4. Calculation: 2004+1/2-4/3+5/2-13+9/2-16/3+...+4005/2-6010.
Disassemble the formula:
Part I: The numerator with denominator 2 has four more terms than the previous term, and they are all positive numbers:1/2+5/2+9/2+13/2 ... 4005/2.
Namely:1/2+(1/2+2)+(1/2+4)+(1/2+6) ...+(1/2+2002) =
The above formula: 1004003
(The * in the above formula stands for multiplication symbol)
The part with denominator 3 is the same.
There should be an easier way. I'll look again.
5. If a.b.c.d is an integer and (a 2+b 2) (c 2+d 2) =1997, then a 2+b 2+c 2+d 2 =1998.
Is this ... is this a proof question or did you forget to write it?
You must have forgotten to write it, otherwise you can't do it.
I'll have a meal first, and I'll cook the rest when I get back.
By the way, which school are you from?
Why don't you ask the teacher? It's almost the end of the term-it's faster to ask the teacher!