Question

chứng minh rằng: \(\sqrt[3]{18-5\sqrt{13}}+\sqrt[3]{18+5\sqrt{13}}\) là các sớ nguyên

Answer 1

Đặt: \(P=\sqrt[3]{18-5\sqrt{13}}+\sqrt[3]{18+5\sqrt{13}}\)

=> \(P^3=18-5\sqrt{13}+18+5\sqrt{13}+3\left(\sqrt[3]{18-5\sqrt{13}}\right)^2.\sqrt[3]{18+5\sqrt{13}}\)\(+3\sqrt[3]{18-5\sqrt{13}}.\left(\sqrt[3]{18+5\sqrt{13}}\right)^2\)

=> \(P^3=36+3\sqrt[3]{18-5\sqrt{13}}.\sqrt[3]{18+5\sqrt{13}}\left(\sqrt[3]{18-5\sqrt{13}}+\sqrt[3]{18+5\sqrt{13}}\right)\)

<=> \(P^3=36+3\sqrt[3]{18^2-25.13}\left(\sqrt[3]{18-5\sqrt{13}}+\sqrt[3]{18+5\sqrt{13}}\right)\)

=> P³=36-3.P

<=> P³+3P-36=0

<=> P³-27 + 3P-9=0

<=> (P-3)(P²+3P+9)+3(P-3)=0

<=> (P-3)(P²+3P+12)=0  

=> P-3=0   (Do P²+3P+12 > 0 với mọi P)

=> P=3

Vậy \(P=\sqrt[3]{18-5\sqrt{13}}+\sqrt[3]{18+5\sqrt{13}}\)= 3 Là 1 số nguyên

Answer 2

Cách khác:

Đặt: \(Q=\sqrt[3]{18-5\sqrt{13}}+\sqrt[3]{18+5\sqrt{13}}\)

\(2Q=\sqrt[3]{144-40\sqrt{13}}+\sqrt[3]{144+40\sqrt{13}}\)

\(=\sqrt[3]{27-27\sqrt{13}+117-13\sqrt{13}}+\sqrt[3]{27+27\sqrt{13}+117+13\sqrt{13}}\)

\(=\sqrt[3]{\left(3-\sqrt{13}\right)^3}+\sqrt[3]{\left(3+\sqrt{13}\right)^3}\)

\(=3-\sqrt{13}+3+\sqrt{13}=6\)

\(\Rightarrow Q=3\)