Question

Solve the given expression:

$\frac{(0\times 1)}{(1\times (-1\left)\right)}+\frac{(-1)\times 19}{(-19)\times 1}+\frac{(-91)\times 0}{(-17)\times 1}+\frac{1\times 19}{(-1)\times 19}$

Answer 1

In the given expression,
$\frac{(0\times 1)}{(1\times (-1\left)\right)}+\frac{(-1)\times 19}{(-19)\times 1}+\frac{(-91)\times 0}{(-17)\times 1}+\frac{1\times 19}{(-1)\times 19}$We know that, multiplication of an integer by one results the same integer and division of an integer by zero results in zero.
So, $\frac{0\times 1}{1\times (-1)}=\frac{0}{-1}=0\frac{(-1)\times 19}{(-19)\times 1}=\frac{-19}{-19}=1\frac{(-91)\times 0}{(-17)\times 1}=\frac{0}{-17}=0\frac{1\times 19}{(-1)\times 19}=\frac{19}{-19}=-1$
Thus, expression
$\frac{(0\times 1)}{(1\times (-1\left)\right)}+\frac{(-1)\times 19}{(-19)\times 1}+\frac{(-91)\times 0}{(-17)\times 1}+\frac{1\times 19}{(-1)\times 19}$$=0+1-0-1$
The answer of the expression is 0.