I assume you want to evaluate the expression. So, let's use the priority rules properly: first of all, parenthesis. So, for now, we concentrate on
[tex] 1 \times \cfrac{2}{7}-\cfrac{1}{3} [/tex]
We again use the priority rules to remember that we need to perform multiplications first, and then additions/subtractions. So, first of all we have to evaluate
[tex] 1 \times \cfrac{2}{7} = \cfrac{2}{7} [/tex]
Since every number multiplied by 1 remains unchanged. The expression in parenthesis is simplified to
[tex] \cfrac{2}{7}-\cfrac{1}{3} [/tex]
Transform the denominators of both fractions to 21 in order to subtract them:
[tex] \cfrac{2}{7}\cdot\cfrac{3}{3}-\cfrac{1}{3}\cdot\cfrac{7}{7} = \cfrac{6}{21}-\cfrac{7}{21} = -\cfrac{1}{21} [/tex].
Now that we simplified the expression in the parenthesis, we can focus on the whole expression again, which has become
[tex] -\cfrac{1}{21} \times 1 \times \cfrac{3}{4} [/tex]
We're left with multiplications only, so we can perform them in any order, since the multiplication is commutative. As before, we can ignore the multiplication by 1, since it doesn't change the value of the expression:
[tex] -\cfrac{1}{21} \times \cfrac{3}{4} [/tex]
Finally, to multiply two fractions, you simply multiply numerators and denominators among them:
[tex] -\cfrac{1}{21} \times \cfrac{3}{4} = -\cfrac{1\cdot3}{21\cdot4} = -\cfrac{3}{84} = -\cfrac{1}{28} [/tex]
