Question

Nếu \(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=\sqrt[3]{a+b+c}\)

Chứng minh rằng : Với mọi số nguyên dương lẻ n , ta đều có : \(\sqrt[n]{a}+\sqrt[n]{b}+\sqrt[n]{c}=\sqrt[n]{a+b+c}\)

Answer 1

\(\left(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\right)^3=a+b+c\)

\(\Leftrightarrow a+b+c+3\left(\sqrt[3]{a}+\sqrt[3]{b}\right)\left(\sqrt[3]{b}+\sqrt[3]{c}\right)\left(\sqrt[3]{a}+\sqrt[3]{c}\right)=a+b+c\)

\(\Leftrightarrow3\left(\sqrt[3]{a}+\sqrt[3]{b}\right)\left(\sqrt[3]{b}+\sqrt[3]{c}\right)\left(\sqrt[3]{a}+\sqrt[3]{c}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b=0\\b+c=0\\c+a=0\end{matrix}\right.\)

+Neu a+b =0 => \(\sqrt[n]{a}+\sqrt[n]{b}=0\)( n : le)=> \(VT=VP=\sqrt[n]{c}\)(dpcm)

Tuong tu cac TH

=> KL