Question

a,tìm số tự nhiên n biết :\(\dfrac{12}{(-2)^{n}}\) =\(\dfrac{-12}{8}\)

b, so sánh \(\dfrac{4}{9}\) và\(\dfrac{5}{8}\) ; -\(\sqrt{\dfrac{4}{9}}\) và\(\dfrac{-3}{4}\)

Answer 1

a) \(\dfrac{12}{\left(-2\right)^n}=\dfrac{-12}{8}\)

\(\Rightarrow12.8=\left(-2\right)^n.\left(-12\right)\)

\(\Rightarrow96=\left(-2\right)^n.\left(-12\right)\)

\(\Rightarrow\left(-2\right)^n=\dfrac{96}{-12}\)

\(\Rightarrow\left(-2\right)^n=-8\)

\(\Rightarrow\left(-2\right)^n=\left(-2\right)^3\)

\(\Rightarrow n=3\)

Vậy \(n=3\)

Answer 2

2)

a) \(\dfrac{4}{9}\) và \(\dfrac{5}{8}\) Mẫu chung: 72

\(\dfrac{4}{9}=\dfrac{4.8}{72}=\dfrac{32}{72}\)

\(\dfrac{5}{8}=\dfrac{5.9}{72}=\dfrac{45}{72}\)

Vì \(\dfrac{32}{72}< \dfrac{45}{72}\)

Vậy \(\dfrac{4}{9}< \dfrac{5}{8}\)

b) \(-\sqrt{\dfrac{4}{9}}\) và \(\dfrac{-3}{4}\) MTC: 12

\(-\sqrt{\dfrac{4}{9}}=-\sqrt{\left(\dfrac{2}{3}\right)^2}=-\dfrac{2}{3}=\dfrac{-2.4}{12}=\dfrac{-8}{12}\)

\(-\dfrac{3}{4}=\dfrac{-3.3}{12}=\dfrac{-9}{12}\)

Vì \(\dfrac{-8}{12}>\dfrac{-9}{12}\)

Vậy \(-\sqrt{\dfrac{4}{9}}>\dfrac{-3}{4}\)