Question Number 165969 by leicianocosta last updated on 10/Feb/22
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Commented by leicianocosta last updated on 10/Feb/22
Commented by hknkrc46 last updated on 10/Feb/22
$$\blacktriangleright\:\left(−\mathrm{3}\right)^{\mathrm{0}} \:=\:\mathrm{1}\:\:\because\:\:\forall\:\boldsymbol{{x}}\:\in\:\mathbb{R}^{\ast} \:\Rightarrow\:\boldsymbol{{x}}^{\mathrm{0}} \:=\:\mathrm{1} \\ $$$$\left(\:\mathbb{R}^{\ast} \:=\:\mathbb{R}\:−\:\left\{\mathrm{0}\right\}\:\right) \\ $$$$\blacktriangleright\:−\left(−\mathrm{3}\right)^{\mathrm{3}} \:=\:−\left(−\mathrm{3}^{\mathrm{3}} \right)\:=\:−\left(−\mathrm{27}\right)\:=\:\mathrm{27} \\ $$$$\:\because\:\:\mathrm{2}\:\nmid\:\boldsymbol{{a}}\:\:\wedge\:\:\boldsymbol{{b}}\:\in\:\mathbb{R}^{+} \:\Rightarrow\:\left(−\boldsymbol{{b}}\right)^{\boldsymbol{{a}}} \:=\:−\boldsymbol{{b}}^{\boldsymbol{{a}}} \\ $$$$\blacktriangleright\:\left(−\mathrm{2}\right)^{−\mathrm{1}} \:=\:\frac{\mathrm{1}}{\left(−\mathrm{2}\right)^{\mathrm{1}} }\:=\:\frac{\mathrm{1}}{−\mathrm{2}^{\mathrm{1}} }\:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\blacktriangleright\:−\left(\mathrm{5}^{−\mathrm{2}} \right)\:=\:−\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }\:=\:−\frac{\mathrm{1}}{\mathrm{25}} \\ $$$$\bigstar\:\boldsymbol{{A}}\:=\:\frac{\mathrm{1}\:\centerdot\:\mathrm{27}\:\centerdot\:\mathrm{5}}{\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(−\frac{\mathrm{1}}{\mathrm{25}}\right)}\:=\:\mathrm{6750} \\ $$
Commented by alephzero last updated on 11/Feb/22
$${A}\:=\:\frac{\left(−\mathrm{3}\right)^{\mathrm{0}} \left(−\left(−\mathrm{3}\right)^{\mathrm{3}} \right)\mathrm{5}}{\left(−\mathrm{2}\right)^{−\mathrm{1}} \left(−\left(\mathrm{5}^{−\mathrm{2}} \right)\right)}\:= \\ $$$$=\:\frac{\mathrm{5}\left(\mathrm{27}\right)}{\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{25}\right)}}\:=\:\mathrm{5}\:\centerdot\:\mathrm{27}\:\centerdot\:\mathrm{2}\:\centerdot\:\mathrm{25}\:=\:\mathrm{270}\:\centerdot\:\mathrm{25} \\ $$$$=\:\mathrm{6750} \\ $$