Question

tập nghiệm của phương trình 8(x²+\(\frac{1}{x^2}\))-34(x+\(\frac{1}{x}\))+51=0 là S={...}

Answer 1

htfjgyjgjghjmhjbhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

Answer 2

I don't know because I in class 6

Answer 3

\(8\left(y^2-2\right)-34y+51=0\)

\(y^2-\frac{34}{8}y+\frac{35}{8}=0\)

\(\left(y-\frac{17}{8}\right)^2=\left(\frac{17}{8}\right)^2-\frac{35}{8}=\frac{9}{8}\)\(\Rightarrow\orbr{\begin{cases}y=\frac{17-6\sqrt{2}}{8}\\y=\frac{17+6\sqrt{2}}{8}\end{cases}}\) Sao đề ra lẻ thế (x+1/x)=y 

Answer 4

\(\orbr{\begin{cases}y=\frac{17}{8}+\frac{3}{8}=\frac{20}{8}=\frac{5}{2}\\y=\frac{17}{8}-\frac{3}{8}=\frac{14}{8}=\frac{7}{4}\left(loai\right)\end{cases}}\)

\(S=\left\{\left(\frac{5-2\sqrt{21}}{4};\frac{5+2\sqrt{21}}{4}\right)\right\}\)