Question Number 204590 by sphelele last updated on 22/Feb/24
Answered by A5T last updated on 22/Feb/24
$$\left.{a}\right)=\frac{\mathrm{2}^{{n}+\mathrm{2}+\mathrm{2}{n}+\mathrm{2}} }{\mathrm{2}^{\mathrm{3}{n}−\mathrm{3}} }=\frac{\mathrm{2}^{\mathrm{3}{n}+\mathrm{4}} }{\mathrm{2}^{\mathrm{3}{n}+\mathrm{4}} \mathrm{2}^{−\mathrm{7}} }=\mathrm{128} \\ $$$$\left.{b}\right)=\left(\frac{{b}−{a}}{{ab}}\overset{−\mathrm{1}} {\right)}=\frac{{ab}}{{b}−{a}} \\ $$$$\left.{c}\right)=\frac{\mathrm{3}^{{m}+\mathrm{1}} \left(\mathrm{3}^{\mathrm{3}} −\mathrm{6}\right)}{\mathrm{7}×\mathrm{3}×\mathrm{3}^{{m}+\mathrm{1}} }=\mathrm{1} \\ $$$$\left({d}\right)=\frac{\sqrt{{p}−{q}\:}\left(\sqrt{{p}+{q}}\right)\left({p}+{q}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} }{\:\sqrt{{p}−{q}}}=\:\:\left({p}+{q}\right)^{\mathrm{3}} ={p}^{\mathrm{3}} +{q}^{\mathrm{3}} +\mathrm{3}{pq}\left({p}+{q}\right) \\ $$