Question Number 158191 by ghakhan88 last updated on 31/Oct/21
$${Can}\:{we}\:{reach}\:{to}\:\frac{{m}}{\left({m}−\mathrm{1}\right){s}}\:+\:\frac{{m}+\mathrm{1}}{{ms}^{\mathrm{2}} }\:{from}\: \\ $$$$ \\ $$$$\:\:\:\:\frac{{ms}+{m}\left({m}+\mathrm{1}\right)}{{s}\left({m}−{s}\right)}\:?? \\ $$
Commented by Rasheed.Sindhi last updated on 01/Nov/21
$$\mathcal{T}{hey}'{re}\:{not}\:{equal}\:{even}! \\ $$$${for}\:{m}=\mathrm{2}\:\&\:{s}=\mathrm{1}\:{they}\:{have}\:{different} \\ $$$${values}. \\ $$$$\:\frac{{m}}{\left({m}−\mathrm{1}\right){s}}\:+\:\frac{{m}+\mathrm{1}}{{ms}^{\mathrm{2}} }\:\overset{?} {=}\frac{{ms}+{m}\left({m}+\mathrm{1}\right)}{{s}\left({m}−{s}\right)} \\ $$$$\:\frac{\mathrm{2}}{\left(\mathrm{2}−\mathrm{1}\right)\left(\mathrm{1}\right)}\:+\:\frac{\mathrm{2}+\mathrm{1}}{\mathrm{2}\left(\mathrm{1}^{\mathrm{2}} \right)}\:\overset{?} {=}\frac{\mathrm{2}\left(\mathrm{1}\right)+\mathrm{2}\left(\mathrm{2}+\mathrm{1}\right)}{\mathrm{1}\left(\mathrm{2}−\mathrm{1}\right)} \\ $$$$\:\frac{\mathrm{2}}{\mathrm{1}}\:+\:\frac{\mathrm{3}}{\mathrm{2}}\overset{?} {\:=}\frac{\mathrm{2}+\mathrm{2}\left(\mathrm{3}\right)}{\mathrm{2}−\mathrm{1}} \\ $$$$\:\:\frac{\mathrm{7}}{\mathrm{2}}\:\overset{?} {=}\frac{\mathrm{2}+\mathrm{6}}{\mathrm{1}} \\ $$$$\frac{\mathrm{7}}{\mathrm{2}}\neq\mathrm{8} \\ $$
Answered by MJS_new last updated on 31/Oct/21
$$\mathrm{no} \\ $$