Tìm hệ số \({x^5}\) trong khai triển Maclaurin của hàm số \(y = \frac{1}{{8 + x}}\)

$\sum _{n=2}^{\infty }\frac{1}{n^2+1}$

$\sum _{n=2}^{\infty }\frac{1}{n\ln n}$

\({\rm{d}}y = \left( {3{x^2} - 18x + 12} \right){\rm{d}}x\)

\({\rm{d}}y = \left( { - 3{x^2} - 18x + 12} \right){\rm{d}}x\)

\({\rm{d}}y = - \left( {3{x^2} - 18x + 12} \right){\rm{d}}x\)

\({\rm{d}}y = \left( { - 3{x^2} + 18x - 12} \right){\rm{d}}x\)

\(dy = 10{(3x + 1)^9}dx\)

\(dy = 30{(3x + 1)^{10}}dx\)

\(dy = 9{(3x + 1)^{10}}dx\)

\(dy = 30{(3x + 1)^9}dx\)

\(dy = \left( {\cos 2x + 3{{\sin }^2}x\cos x} \right)dx\)

\(dy = \left( {2\cos 2x + 3{{\sin }^2}x\cos x} \right)dx\)

\(dy = \left( {2\cos 2x + {{\sin }^2}x\cos x} \right)dx\)

\(dy = \left( {\cos 2x + {{\sin }^2}x\cos x} \right)dx\)

\(dy = \frac{1}{{\sqrt[3]{{{{(x + 1)}^2}}}}}dx\)

\(dy = \frac{3}{{\sqrt[3]{{{{(x + 1)}^2}}}}}dx\)

\(dy = \frac{2}{{\sqrt[3]{{{{(x + 1)}^2}}}}}dx\)

\(dy = \frac{1}{{3\sqrt[3]{{{{(x + 1)}^2}}}}}dx\)

\({\rm{d}}y = \left( {3{x^2} - 5} \right){\rm{d}}x\)

\({\rm{d}}y = - \left( {3{x^2} - 5} \right){\rm{d}}x\)

\({\rm{d}}y = \left( {3{x^2} + 5} \right){\rm{d}}x\)

\({\rm{d}}y = \left( {3{x^2} - 5} \right){\rm{d}}x\)

\({\rm{d}}y = \frac{1}{4}{\rm{d}}x\)

\({\rm{d}}y = \frac{1}{{{x^4}}}{\rm{d}}x\)

\({\rm{d}}y = - \frac{1}{{{x^4}}}{\rm{d}}x\)

\({\rm{d}}y = {x^4}{\rm{d}}x\)