Cho hàm \[{\rm{z}} = {{\rm{x}}^2} - {\rm{y}} - \ln |{\rm{y}}| - 2\]. Khẳng định nào sau đây đúng?

A. 1

B. 2

C. 0

D. \(\frac{1}{2}\)

A. \[{{\rm{d}}^{\rm{2}}}{\rm{z = 2cos2xd}}{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{e}}^{{{\rm{y}}^{\rm{2}}}}}{\rm{(4}}{{\rm{y}}^{\rm{2}}}{\rm{ + 2)d}}{{\rm{y}}^{\rm{2}}}\]

B. \[{{\rm{d}}^{\rm{2}}}{\rm{z = 2cos2xd}}{{\rm{x}}^{\rm{2}}}{\rm{ + 2}}{{\rm{e}}^{{{\rm{y}}^{\rm{2}}}}}{\rm{d}}{{\rm{y}}^{\rm{2}}}\]

C. \[{{\rm{d}}^{\rm{2}}}{\rm{z = }} - {\rm{2cos2xd}}{{\rm{x}}^{\rm{2}}}{\rm{ + 2y}}{{\rm{e}}^{{{\rm{y}}^{\rm{2}}}}}{\rm{d}}{{\rm{y}}^{\rm{2}}}\]

D. \[{{\rm{d}}^{\rm{2}}}{\rm{z = cos2xd}}{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{e}}^{{{\rm{y}}^{\rm{2}}}}}{\rm{d}}{{\rm{y}}^{\rm{2}}}\]

A. \[ - \frac{{\rm{x}}}{{{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}}}\]

B. \[\frac{{\rm{x}}}{{{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}}}\]

C. \[ - \frac{{\rm{y}}}{{{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}}}\]

D. \[ - \frac{{\rm{1}}}{{{{\rm{x}}^{\rm{2}}}{\rm{y + }}{{\rm{y}}^{\rm{3}}}}}\]

A.  Nếu \[{{\rm{u}}_{\rm{n}}} \to 0\,\,{\rm{khi}}\,\,\,\,{\rm{n}} \to \infty \]thì chuỗi trên hội tụ

B. Nếu \[{{\rm{u}}_{\rm{n}}} \to 0\,\,{\rm{khi}}\,\,\,\,{\rm{n}} \to \infty \] thì chuỗi trên phân kỳ

C. Nếu chuỗi trên phân kỳ thì\[{{\rm{u}}_{\rm{n}}} \to 0\,\,{\rm{khi}}\,\,\,\,{\rm{n}} \to \infty \]

D. Nếu chuỗi trên hội tụ thì \[{{\rm{u}}_{\rm{n}}} \to 0\,\,{\rm{khi}}\,\,\,\,{\rm{n}} \to \infty \]

A. 1

B. \[\frac{1}{{60}}\]

C. \[\frac{1}{{300}}\]

D. \[\frac{2}{{150}}\]

A. 0

B. 1

C. \[\frac{1}{2}\]

D. \[ - \frac{1}{2}\]

A. 0

B. 1

C. \(\frac{1}{2}\)

D. \( - \frac{1}{2}\)