Question

14. Compare os radicais abaixo e determine qual é o maior deles.​

Answer 1

$\boxed{\begin{array}{l}\tt a)~\sf\sqrt{21}<\sqrt{48}\\\tt b)~\begin{cases}\sf \sqrt{12}=\sqrt[\sf6]{\sf 12^3}\\\sf \sqrt[\sf3]{\sf 31}=\sqrt[\sf6]{\sf31^2}\end{cases}\\\sf 12<31(I)\\\sf 12^2<31^2(II)\\\sf 12^2<12^3(III)\\\sf substituindo~a~desigualde~II~em~III~temos:\\\sf31^2<12^3\\\sf portanto~\sqrt[\sf6]{\sf 12^3}>\sqrt[\sf6]{\sf 31^2}\longrightarrow \sqrt{12}>\sqrt[\sf3]{\sf31}\end{array}}$

$\boxed{\begin{array}{l}\tt c)~\begin{cases}\sf\sqrt[\sf3]{\sf24}=\sqrt[\sf12]{\sf24^4}\\\sf\sqrt[\sf4]{\sf80}=\sqrt[\sf12]{\sf80^3}\end{cases}\\\sf 24<80~(I)\\\sf 24^3<80^3~(II)\\\sf 24^3<24^4~(III)\\\sf substituindo~a~desigualdade~II~em~III~temos:\\\sf 80^3<24^4\\\sf ou~seja\\\sf \sqrt[\sf12]{\sf24^4}>\sqrt[\sf12]{\sf80^3}\longrightarrow \sqrt[\sf3]{\sf24}>\sqrt[\sf4]{\sf80}\end{array}}$

$\large\boxed{\begin{array}{l}\tt d)\sf \sqrt{36}=\sqrt[\sf4]{\sf 36^2}\\\sf\sqrt[\sf4]{\sf 256}=\sqrt[\sf4]{\sf2^8}\\\sf2<36(I)\\\sf 2^2<36^2(II)\\\sf 2^2<2^8(III)\\\sf 2^8<36^2(IV)\\\sf isto~\acute e\\\sf \sqrt[\sf4]{\sf36^2}>\sqrt[\sf4]{\sf2^8}\longrightarrow\sqrt{36}>\sqrt[\sf4]{\sf256}\end{array}}$

$\large\boxed{\begin{array}{l}\tt e)~\begin{cases}\sqrt[\sf3]{\sf7^2}=\sqrt[\sf6]{\sf7^4}\\\sf\sqrt{9^5}=\sqrt[\sf6]{\sf9^{15}}\end{cases}\\\sf7<9\\\sf7^4<9^4\\\sf9^4<9^{15}\\\sf 7^4<9^{15}\\\sf ou~seja\\\sf\sqrt[\sf6]{\sf7^4}<\sqrt[\sf6]{\sf9^{15}}\longrightarrow\sqrt[\sf3]{\sf7^2}<\sqrt{9^5}\end{array}}$

$\large\boxed{\begin{array}{l}\tt f)\begin{cases}\sf\sqrt[\sf4]{\sf4^3}=\sqrt[\sf20]{\sf4^{15}}\\\sf\sqrt[\sf5]{\sf7^4}=\sqrt[\sf20]{\sf7^{16}}\end{cases}\\\sf 4<7\\\sf 4^{15}<7^{15}\\\sf 7^{15}<7^{16}\\\sf 4^{15}<7^{16}\\\sf ou~seja\\\sf \sqrt[\sf20]{\sf4^{15}}<\sqrt[\sf20]{\sf7^{16}}\longrightarrow\sqrt[\sf4]{\sf4^3}<\sqrt[\sf5]{\sf7^4}\end{array}}$

$\large\boxed{\begin{array}{l}\tt g)~\begin{cases}\sf\sqrt[\sf6]{\sf2^8}=\sqrt[\sf12]{\sf6^{16}}\\\sf\sqrt[\sf4]{\sf4^5}=\sqrt[\sf12]{\sf4^{15}}\end{cases}\\\sf 4<6\\\sf 4^{15}<6^{15}\\\sf 6^{15}<6^{16}\\\sf 4^{15}<6^{16}\\\sf portanto\\\sf \sqrt[\sf12]{\sf6^{16}}>\sqrt[\sf12]{\sf4^{15}}\longrightarrow \sqrt[\sf6]{\sf2^8}>\sqrt[\sf4]{\sf4^5}\end{array}}$

$\large\boxed{\begin{array}{l}\tt h)\begin{cases}\sf\sqrt[\sf3]{\sf16}=\sqrt[\sf3]{\sf2^4}=\sqrt[\sf12]{\sf2^{16}}\\\sf\sqrt[\sf4]{\sf4^6}=\sqrt[\sf4]{\sf2^{12}}=\sqrt[\sf12]{\sf2^{36}}\end{cases}\\\sf 2^{16}<2^{36}\\\sf portanto\\\sf \sqrt[\sf12]{\sf2^{16}}<\sqrt[\sf12]{\sf2^{36}}\longrightarrow \sqrt[\sf3]{\sf16}<\sqrt[\sf4]{\sf4^6}\end{array}}$