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Question-165969




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
▶ (−3)^0  = 1  ∵  ∀ x ∈ R^∗  ⇒ x^0  = 1  ( R^∗  = R − {0} )  ▶ −(−3)^3  = −(−3^3 ) = −(−27) = 27   ∵  2 ∤ a  ∧  b ∈ R^+  ⇒ (−b)^a  = −b^a   ▶ (−2)^(−1)  = (1/((−2)^1 )) = (1/(−2^1 )) = −(1/2)  ▶ −(5^(−2) ) = −(1/5^2 ) = −(1/(25))  ★ A = ((1 ∙ 27 ∙ 5)/((−(1/2))(−(1/(25))))) = 6750
$$\blacktriangleright\:\left(−\mathrm{3}\right)^{\mathrm{0}} \:=\:\mathrm{1}\:\:\because\:\:\forall\:\boldsymbol{{x}}\:\in\:\mathbb{R}^{\ast} \:\Rightarrow\:\boldsymbol{{x}}^{\mathrm{0}} \:=\:\mathrm{1} \\ $$$$\left(\:\mathbb{R}^{\ast} \:=\:\mathbb{R}\:−\:\left\{\mathrm{0}\right\}\:\right) \\ $$$$\blacktriangleright\:−\left(−\mathrm{3}\right)^{\mathrm{3}} \:=\:−\left(−\mathrm{3}^{\mathrm{3}} \right)\:=\:−\left(−\mathrm{27}\right)\:=\:\mathrm{27} \\ $$$$\:\because\:\:\mathrm{2}\:\nmid\:\boldsymbol{{a}}\:\:\wedge\:\:\boldsymbol{{b}}\:\in\:\mathbb{R}^{+} \:\Rightarrow\:\left(−\boldsymbol{{b}}\right)^{\boldsymbol{{a}}} \:=\:−\boldsymbol{{b}}^{\boldsymbol{{a}}} \\ $$$$\blacktriangleright\:\left(−\mathrm{2}\right)^{−\mathrm{1}} \:=\:\frac{\mathrm{1}}{\left(−\mathrm{2}\right)^{\mathrm{1}} }\:=\:\frac{\mathrm{1}}{−\mathrm{2}^{\mathrm{1}} }\:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\blacktriangleright\:−\left(\mathrm{5}^{−\mathrm{2}} \right)\:=\:−\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }\:=\:−\frac{\mathrm{1}}{\mathrm{25}} \\ $$$$\bigstar\:\boldsymbol{{A}}\:=\:\frac{\mathrm{1}\:\centerdot\:\mathrm{27}\:\centerdot\:\mathrm{5}}{\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(−\frac{\mathrm{1}}{\mathrm{25}}\right)}\:=\:\mathrm{6750} \\ $$
Commented by alephzero last updated on 11/Feb/22
A = (((−3)^0 (−(−3)^3 )5)/((−2)^(−1) (−(5^(−2) )))) =  = ((5(27))/(1/(2(25)))) = 5 ∙ 27 ∙ 2 ∙ 25 = 270 ∙ 25  = 6750
$${A}\:=\:\frac{\left(−\mathrm{3}\right)^{\mathrm{0}} \left(−\left(−\mathrm{3}\right)^{\mathrm{3}} \right)\mathrm{5}}{\left(−\mathrm{2}\right)^{−\mathrm{1}} \left(−\left(\mathrm{5}^{−\mathrm{2}} \right)\right)}\:= \\ $$$$=\:\frac{\mathrm{5}\left(\mathrm{27}\right)}{\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{25}\right)}}\:=\:\mathrm{5}\:\centerdot\:\mathrm{27}\:\centerdot\:\mathrm{2}\:\centerdot\:\mathrm{25}\:=\:\mathrm{270}\:\centerdot\:\mathrm{25} \\ $$$$=\:\mathrm{6750} \\ $$

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