If 3b 3 2a b 3 0, find the value of 5 2a b 2 6a 2b 2 4a 3b to the process

If | 3b-3|+|2a-b-3 |=0, find the value of 5(2a-b)-2(6a-2b+2)+(4a-3b) to the process.

Solution: |3b-3|AND |2a-b-3 |are two non-negative numbers, the sum of which is 0, and there must be:

3b-3=0...1)

2a-b-3=0...2)

b = 1 from (1) and a = 2 by (2).

Therefore, 5(2a-b)-2(6a-2b+2)+(4a-3b)=(10a-5b)-(12a-4b+4)+(4a-3b).

2a-4b-4=2×2-4×1-4=-4

Because the absolute value is greater than or equal to 0, the sum of the two absolute values is equal to 0, so 3b-3=0, b=1, 2a-b-3=0, a=2

Substitute a=2,b=1 into 5(2a-b)-2(6a-2b+2)+(4a-3b) to get -4

It should be 0, not -4, how do you calculate it, 2a-4b is 2x2-4x1, and the result is 0

Because the absolute value is not negative.

So 3b-2=0 2a-b-3=0

b=2/3 2a-2/3-3=0

a=11/6

Substitute the values of a and b into 5(2a-b)-2(6a-2b+2)+(4a-3b-1 2).

Because the absolute value is a non-negative number.

So |3b-2|=0 |2a-b-3|=0 dang|3b-2|= 0, b = two-thirds.

When|2a-b-3|=0, b = two-thirds, a = eleven sixths.

After a and b are calculated, simplify the formula first, and then substitute it to calculate. (I've just learned, I don't know if it's right).

Known |3a+b+5|+|2a-2b-2|=0 because |3a+b+5|>=0,|2a-2b-2|>=0, so |3a+b+5|=0,|2a-2b-2|=0,3a+b+5=0,..1)

2a-2b-2=0,..2)

6a+2b+10+2a-2b-2=0,8a=-8,a=-1,a=-1 substituted into 2):

2(-1)-2b-2=0,b=-2;

2a²-3b=2(-1)²-3(-2)=2+6=8;

Solution: 2a+3b) -a+5b)(a-5b)+(2a-3b) 2(4a+12ab+9b)-a-25b)+4a-12ab+9b)

4a +12ab+9b -a +25b +4a -12ab+9b

7a²-43b²

4a^2-b^2+3(4a^2-4ab+b^2)+(9a^2+12ab)-2a^2+3

4a^2-b^2+12a^2-12ab+3b^2-9a^2+12ab-2a^2+3

5a^2+2b^2+3

When the absolute width of the segment is a=-1, b=1 2, the grip is bright.

Original = 5(-1) 2+2(1 2) 2+317 macro pre-2

(2a+3b)(2a-3b)+(a-3b)^2=4a²-9b²+a²-6ab+9b²..

5a²-6ab...Merge items of the same kind.

a(5a-6b) .Extract the common factor.