|x|+|x+ 1|= 1

It means that the sum of the distances from any point X to two points x=0 and x=- 1 on the number axis is 1.

Obviously, the distance between x=0 and x=- 1 is 1.

Then, x is any point between x=0 and x=- 1.

That is-1≤x≤0.

2、

|z|= 1, let z = cosθ+isθ.

Then | z+2 √ 2+I | =| (cos θ+2 √ 2)+(1+sin θ) I |

=√[(cosθ+2√2)？ +( 1+sinθ)？ ]

=√(cos？ θ+8+4√2cosθ+ 1+2sinθ+sin？ θ)

=√( 10+4√2cosθ+2sinθ)

=√[ 10+6sin(θ+φ)]

Then the maximum value is √( 10+6)=4.

3、

Let the term number of arithmetic progression be 2n+ 1.

Then, odd terms have n+ 1 terms, and even terms have n terms. The middle term is a.

And the sum of odd terms = (n/2) * [2 * a.

Sum of even terms = (n/2) * [2 * a

Subtract the two expressions to get

The substitution of (1) is: (n+ 1)*29=290.

Therefore, n+ 1= 10.

Then, n=9

Therefore, the number of terms =2n+ 1= 19.

-answer: c