\(\Leftrightarrow\left(m^2-5\right)x=m+1\)

\(\Leftrightarrow x=\frac{m+1}{m^2-5}\)

Neu \(m=\sqrt{5}\)thi PT vo nghiem

Goij số hạt notron, proton, electron của nguyên tử Y lần lượt là N, P, E trong đó P = E

=> N + P + E =58

=> N + 2 E =58 (1)

Mặt khác: (N + P ) - ( E + P) =1

=> N - E = 1  (2)

Từ (1); (2) => N = 20 , E = P =19 

=> Y là nguyên tử K 

\(DK:x\le\sqrt[8]{17}\)

\(\Leftrightarrow\left(\sqrt[4]{17-x^8}-2\right)+\left(1-\sqrt[3]{2x^8-1}\right)=0\)

\(\Leftrightarrow\frac{\sqrt{17-x^8}-4}{\sqrt[4]{17-x^8}+2}+\frac{2\left(1-x^8\right)}{1+\sqrt[3]{2x^8-1}+\left(\sqrt[3]{2x^8-1}\right)^2}=0\)

\(\Leftrightarrow\frac{1-x^8}{\left(\sqrt[4]{17-x^8}+2\right)\left(\sqrt{17-x^8}+4\right)}+\frac{2\left(1-x^8\right)}{1+\sqrt[3]{2x^8-1}+\left(\sqrt[3]{2x^8-1}\right)^2}=0\)

\(\Leftrightarrow\left(1-x^8\right)\left[\frac{1}{\left(\sqrt[4]{17-x^8}+2\right)\left(\sqrt{17-x^8}\right)}+\frac{1}{1+\sqrt[3]{2x^8-1}+\left(\sqrt[3]{2x^8-1}\right)}\right]=0\)

Vi \(\frac{1}{\left(\sqrt[4]{17-x^8}+2\right)\left(\sqrt{17-x^8}\right)}+\frac{2}{1+\sqrt[3]{2x^8-1}+\left(\sqrt[3]{2x^8-1}\right)^8}>0\left(\forall x\le\sqrt[8]{17}\right)\)

\(\Rightarrow x^8=1\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\left(l\right)\\x=-1\left(n\right)\end{cases}}\)

Vay nghiem cua PT la \(x=-1\)