Question

$\frac{3}{7}+(\frac{-6}{11})+(\frac{-8}{21})+(\frac{5}{22})$ $=(\frac{-125}{462})$. If true enter $1$ else $0$.

Answer 1

L.H.S $=\frac{3}{7}+\left(\frac{-6}{11}\right)+\left(\frac{-8}{21}\right)+\left(\frac{5}{22}\right)$

$=\frac{3}{7}-\frac{6}{11}-\frac{8}{21}+\frac{5}{22}$

$=(\frac{3}{7}-\frac{8}{21})+(\frac{5}{22}-\frac{6}{11})$

$=(\frac{3\times 3}{7\times 3}-\frac{8}{21})+(\frac{5}{22}-\frac{6\times 2}{11\times 2})$

$=(\frac{9}{21}-\frac{8}{21})+(\frac{5}{22}-\frac{12}{22})$

$=\left(\frac{9-8}{21}\right)+\left(\frac{5-12}{22}\right)$

$=\left(\frac{1}{21}\right)+\left(\frac{-7}{22}\right)$

$=\left(\frac{1\times 22}{21\times 22}\right)+\left(\frac{-7\times 21}{22\times 21}\right)$

$=\left(\frac{22}{462}\right)+\left(\frac{-147}{462}\right)$

$=\left(\frac{22-147}{462}\right)$

$=\left(\frac{-125}{462}\right)$

$=$ R.H.S

$\therefore$ It is true the answer is $1$