Bài tập 5.63 trang 209 SBT Toán 11

\(\begin{array}{l}
y = \frac{x}{{{{(1 - x)}^2}{{(1 + x)}^3}}}\\
 \Rightarrow y\prime  = \frac{{{{(1 - x)}^2}{{(1 + x)}^3} - x[ - 2(1 - x){{(1 + x)}^3} + {{(1 - x)}^2}.3{{(1 + x)}^2}]}}{{{{(1 - x)}^4}{{(1 + x)}^6}}}\\
 = \frac{{{{(1 - x)}^2}{{(1 + x)}^3} - x(1 - {x^2})[ - 2(1 + 2x + {x^2}) + 3(1 - {x^2})]}}{{{{(1 - x)}^4}{{(1 + x)}^6}}}\\
 = \frac{{{{(1 - x)}^2}{{(1 + x)}^3} - x(1 - {x^2})( - 5{x^2} - 4x + 1)}}{{{{(1 - x)}^4}{{(1 + x)}^6}}}\\
 = \frac{{{{(1 - x)}^2}{{(1 + x)}^3} - x(1 - {x^2})(1 + x)(1 - 5x)}}{{{{(1 - x)}^4}{{(1 + x)}^6}}}\\
 = \frac{{(1 - x){{(1 + x)}^2}[1 - {x^2} - x(1 - 5x)]}}{{{{(1 - x)}^4}{{(1 + x)}^6}}}\\
 = \frac{{4{x^2} - x + 1}}{{{{(1 - x)}^4}{{(1 + x)}^6}}}
\end{array}\)

-- Mod Toán 11 HỌC247