The number of real solutions of the equation x(x2+3|x|+5|x−1|+6|x−2|)=0 is_______.

The given equation is x(x^2 + 3|x| + 5|x-1| + 6|x-2|) = 0.

This equation will have real solutions when either x = 0 or when the expression in the parentheses equals zero.

1. When x = 0, the equation is satisfied, so x = 0 is one real solution.

2. For the expression in the parentheses to equal zero, we need to consider the absolute value expressions separately for positive and negative values of x.

   a. When x > 0, the equation becomes x(x^2 + 3x + 5(x-1) + 6(x-2)) = 0. Simplifying this gives x(x^2 + 3x + 5x - 5 + 6x - 12) = 0, which simplifies further to x(x^2 + 14x - 17) = 0. This quadratic equation has two real solutions, but since we assumed x > 0, only the positive solution is valid.

   b. When x < 0, the equation becomes x(x^2 - 3x - 5(x-1) - 6(x-2)) = 0. Simplifying this gives x(x^2 - 3x - 5x + 5 - 6x + 12) = 0, which simplifies further to x(x^2 - 14x + 17) = 0. This quadratic equation also has two real solutions, but since we assumed x < 0, only the negative solution is valid.

So, in total, the equation has 1 (from x = 0) + 1 (from x > 0) + 1 (from x < 0) = 3 real solutions.