In arithmetic and algebra, there is a set of rules that need to be used dealing with the orders in which expressions are solved.
This is the BODMAS abbreviation and it stands for Brackets, orders(dealing with roots), divison, multiplication, addition and subtraction. Each expression must be solved in accordance to how these set of operations are arranged.
With reference to the question at hand, we shall be solving the terms in the brackets first, before proceeding to solve the terns which are outside the brackets. We proceed as follows;
a. 7+(9-3^2)
3^2 = 9
7+(9-3^2) = 7 + (9-9) = 7 + (0) = 7 + 0 = 7
b. 6× 1/6+ 5(12/4-3)
starting with the bracket;
12/4 = 3
6× 1/6+ 5(12/4-3) = 6× 1/6+ 5(3-3)
6× 1/6+ 5(0)
6× 1/6+ 5 × 0
we move on to the multiplications;
6 * 1/6 = 1 and 5 * 0 = 0
so we have 1 + 0 = 1
c. (1/5)5×14
This can be written as;
1/5 * 5 * 14 = 1 * 14 = 14
d. [3/2×1] 2/3 = (3/2) 2/3 = 3/2 * 2/3 = 1