\(\frac{4.\left(\frac{1}{5}-\frac{1}{71}+\frac{1}{123}\right)}{3.\left(\frac{1}{5}-\frac{1}{71}+\frac{1}{123}\right)}-\frac{\frac{1}{19}+\frac{1}{1315}-\frac{1}{287}}{3.\left(\frac{1}{19}+\frac{1}{1315}-\frac{1}{287}\right)}=\frac{4}{3}-\frac{1}{3}=\frac{3}{3}=1\)