Có \(ab=\sqrt[3]{\left(7+\sqrt{50}\right)\left(7-\sqrt{50}\right)}\)\(=\sqrt[3]{7^2-\left(\sqrt{50}\right)^2}=-1\)
\(a^3+b^3=7+\sqrt{50}+7-\sqrt{50}=14\)
Lại có: \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)
\(\Leftrightarrow14=M^3+3M\) (do a+b=M)
\(\Leftrightarrow M^3+3M-14=0\)
\(\Leftrightarrow M^3-2M^2+2M^2-4M+7M-14=0\)
\(\Leftrightarrow M^2\left(M-2\right)+2M\left(M-2\right)+7\left(M-2\right)=0\)
\(\Leftrightarrow\left(M-2\right)\left(M^2+2M+7\right)=0\)
\(\Leftrightarrow M=2\) (do \(M^2+2M+7=\left(M+1\right)^2+6>0\forall M\))
=> M là số chẵn
\(N=a^7+b^7=\left(a^6+b^6\right)\left(a+b\right)-ab\left(a^5+b^5\right)\)
\(=2\left(a^6+b^6\right)-\left(-1\right)\left(a^5+b^5\right)\)
\(=2\left(a^6+b^6\right)+\left(a^2+b^2\right)\left(a^3+b^3\right)-a^2b^2\left(a+b\right)\)
\(=2\left(a^6+b^6\right)+14\left(a^2+b^2\right)-2a^2b^2\)
\(=2\left(a^6+b^6+7a^2+7b^2-a^2b^2\right)⋮2\)
=>N là số chẵn
M=a+b=\(\sqrt[3]{7+\sqrt{50}}+\sqrt[3]{7-\sqrt{50}}\)
=>\(M^3=\left(\sqrt[3]{7+\sqrt{50}}+\sqrt[3]{7-\sqrt{50}}\right)^3\)
\(M^3=7+\sqrt{50}+3.\sqrt[3]{7+\sqrt{50}}.(\sqrt[3]{7-\sqrt{50}})^2+3.\left(\sqrt[3]{7+\sqrt{50}}\right)^2.\sqrt[3]{7-\sqrt{50}}+7-\sqrt{50}\)
\(M^3=14+3.\left(\sqrt[3]{7+\sqrt{50}}+\sqrt[3]{7-\sqrt{50}}\right)\sqrt[3]{\left(7-\sqrt{50}\right).(7+\sqrt{50})}\)
\(M^3=\)\(3M\).\(\sqrt[3]{7^2-50}+14\)
\(M^3=-3M+14\)
\(M^3+3M-14=0\)
\(M^3-2M^2+2M^2-4M+7M-14=0\)
<=>\(M^2\left(M-2\right)+2M\left(M-2\right)+7\left(M-2\right)=0< =>\left(M^2+2M+7\right)\left(M-2\right)=0\)
=>M-2=0<=>M=2=>a+b chẵn
giúp em với, em đang cần bài này ạ
