Question

[NGuyễn đình duy hưng]

so sánh 

a) (-27) mũ 27 và (-243)mũ 13 

b) (-1/8)mũ 25 và (-1/128) mũ 13 

c) 4mũ 50 và 8mũ 30 

d) ( 1/9) mũ 17 và (1/27) mũ 12 

dấu / là phần nha như là 1 phần 8 tui ghi là 1/8 nha 

Answer 1

[Nguyễn Đức Trí]

a) Ta có :

\(27^{27}>27^{26}=\left(27^2\right)^{13}=729^{13}>243^{13}\)

\(\Rightarrow27^{27}>243^{13}\)

\(\Rightarrow-27^{27}< -243^{13}\)

\(\Rightarrow\left(-27\right)^{27}< \left(-243\right)^{13}\)

b) \(\left(\dfrac{1}{8}\right)^{25}>\left(\dfrac{1}{8}\right)^{26}=\left(\dfrac{1}{8^2}\right)^{13}=\left(\dfrac{1}{64}\right)^{13}>\left(\dfrac{1}{128}\right)^{13}\)

\(\Rightarrow\left(\dfrac{1}{8}\right)^{25}>\left(\dfrac{1}{128}\right)^{13}\)

\(\Rightarrow\left(-\dfrac{1}{8}\right)^{25}< \left(-\dfrac{1}{128}\right)^{13}\)

c) \(4^{50}=\left(4^5\right)^{10}=1024^{10}\)

\(8^{30}=\left(8^3\right)^{10}=512^{10}< 1024^{10}\)

\(\Rightarrow4^{50}>8^{30}\)

d) \(\left(\dfrac{1}{9}\right)^{17}< \left(\dfrac{1}{9}\right)^{12}< \left(\dfrac{1}{27}\right)^{12}\)

\(\Rightarrow\left(\dfrac{1}{9}\right)^{17}< \left(\dfrac{1}{27}\right)^{12}\)

Answer 2

[Vũ Tuệ Lâm]

a) Ta có :

$27^{27}>27^{26}={\left(27^2\right)}^{13}=729^{13}>243^{13}$

$\Rightarrow 27^{27}>243^{13}$

$\Rightarrow -27^{27}<-243^{13}$

$\Rightarrow {\left(-27\right)}^{27}<{\left(-243\right)}^{13}$

b) ${\left(\genfrac{}{}{0ex}{}{1}{8}\right)}^{25}>{\left(\genfrac{}{}{0ex}{}{1}{8}\right)}^{26}={\left(\genfrac{}{}{0ex}{}{1}{8^2}\right)}^{13}={\left(\genfrac{}{}{0ex}{}{1}{64}\right)}^{13}>{\left(\genfrac{}{}{0ex}{}{1}{128}\right)}^{13}$

$\Rightarrow {\left(\genfrac{}{}{0ex}{}{1}{8}\right)}^{25}>{\left(\genfrac{}{}{0ex}{}{1}{128}\right)}^{13}$

$\Rightarrow {\left(-\genfrac{}{}{0ex}{}{1}{8}\right)}^{25}<{\left(-\genfrac{}{}{0ex}{}{1}{128}\right)}^{13}$

c) $4^{50}={\left(4^5\right)}^{10}=1024^{10}$

$8^{30}={\left(8^3\right)}^{10}=512^{10}<1024^{10}$

$\Rightarrow 4^{50}>8^{30}$

d) ${\left(\genfrac{}{}{0ex}{}{1}{9}\right)}^{17}<{\left(\genfrac{}{}{0ex}{}{1}{9}\right)}^{12}<{\left(\genfrac{}{}{0ex}{}{1}{27}\right)}^{12}$

$\Rightarrow {\left(\genfrac{}{}{0ex}{}{1}{9}\right)}^{17}<{\left(\genfrac{}{}{0ex}{}{1}{27}\right)}^{12}$