The absolute value of 2a 1 5b 4 0 finds the value of ab

2a-1|+5b-4=0 We can know |2a-1|and 5b-4 are opposites of each other, and since the absolute value is non-negative.

So |2a-1|=4-5b

At this time, we will discuss it in two categories.

1）|2a-1|True or non-negative then 2a-1=5b-4==》a=(5b-3) 2 a+b=(7b-3) 2

2) |2a-1|is negative, then 1-2a=5b-4===>a=(5-5b) 2 a+b=(5-3b) 2

The question goes like this::

2a-1=0;

then a=1 2;

5b-4=0;

then b=4 5;

This is my understanding, I don't know if it's right. That's right, thank you.

The question is wrong, this is at most a range partner, for example, there are two groups: a=0, b=3 5; a=1,b=3/5;It's obviously not worth it.

By: |a-2|=5 Available:

a-2=5 or a-2=-5

A=7 or a=-3

In the same way, by |b+1|=4 available:

b=3 or Bi Seeheng b=-5

Then the value of a+b could be the following four shouting values:

Hand made -8

a－6|+|b－5|=0

a－6|0 and |b－5|≥0

The wild number argues that a 6=0 and b 5=0

then a=6, b=5

2a Preamble b = 2 6 5 = 7

Because the absolute value of a is 5 so a = 5 or -5 and the absolute value of b-1 is 2 so b = 3 or -1, so a + b = 8 or -2 or -6 or 4

Solution: According to the problem, a-1=0 b+4=0

Therefore, a=1 b=-4 the original formula should have (2a+b) square the original formula instead of opening brackets to get [4a 2+4ab+b 2-4a 2+b 2-6b] 2b

4ab-6b+b^2]/2b

2a-3+b/2

3 Glad to answer for you, I wish you progress in your studies! If you don't understand, you can ask!

Glad to answer for you :

a-1|+（b+4）^2=0

a-1|=0

a=1（b+4）^2=0

b = -4 brings a b into [(2a+b)-(2a+b)(2a-b)-6b] 2b

to calculate.