Question Number 114134 by sherzodbek last updated on 17/Sep/20
$$\left({m}^{\mathrm{2}} −{n}^{\mathrm{2}} +\mathrm{6}\left({n}+{m}\right)/\left({m}^{\mathrm{2}} −\left(\mathrm{6}−{n}\right)^{\mathrm{2}} \right.\right. \\ $$$${m}+{n}=\mathrm{12} \\ $$
Commented by sherzodbek last updated on 17/Sep/20
yes
Commented by bemath last updated on 17/Sep/20
$${do}\:{you}\:{meant}\:\frac{{m}^{\mathrm{2}} −{n}^{\mathrm{2}} +\mathrm{6}\left({n}+{m}\right)}{{m}^{\mathrm{2}} −\left(\mathrm{6}−{n}\right)^{\mathrm{2}} }\:? \\ $$$${where}\:{m}+{n}\:=\:\mathrm{12} \\ $$
Commented by bemath last updated on 17/Sep/20
$$\Leftrightarrow\:\frac{\mathrm{12}\left({m}−{n}\right)+\mathrm{72}}{\left({m}+\mathrm{6}−{n}\right)\left({m}+{n}−\mathrm{6}\right)}\:= \\ $$$$\frac{\mathrm{12}\left({m}−{n}\right)+\mathrm{72}}{\left({m}−{n}+\mathrm{6}\right)\left(\mathrm{6}\right)}\:=\frac{\mathrm{2}\left({m}−{n}\right)+\mathrm{12}}{{m}−{n}+\mathrm{6}} \\ $$$$=\:\frac{\mathrm{2}\left(\mathrm{12}−{n}−{n}\right)+\mathrm{12}}{\mathrm{12}−{n}−{n}+\mathrm{6}} \\ $$$$=\:\frac{\mathrm{36}−\mathrm{4}{n}}{\mathrm{18}−\mathrm{2}{n}}\:=\:\frac{\mathrm{2}\left(\mathrm{18}−\mathrm{2}{n}\right)}{\mathrm{18}−\mathrm{2}{n}}\:=\:\mathrm{2} \\ $$
Commented by MJS_new last updated on 17/Sep/20
$$\frac{{m}^{\mathrm{2}} −{n}^{\mathrm{2}} +\mathrm{6}\left({n}+{m}\right)}{{m}^{\mathrm{2}} −\left(\mathrm{6}−{n}\right)^{\mathrm{2}} }=\frac{{m}+{n}}{{m}+{n}−\mathrm{6}}=\mathrm{2} \\ $$
Answered by sherzodbek last updated on 17/Sep/20
Commented by mathmax by abdo last updated on 18/Sep/20
$$\mathrm{really}\:\mathrm{good}\:\mathrm{for}\:\mathrm{somebody}.... \\ $$