Giải bài 5.63 - 5.64 trang 209 - SBT Đại số và Giải tích lớp 11

5.63

\(\begin{aligned} & y=\dfrac{x}{{{\left( 1-x \right)}^{2}}{{\left( 1+x \right)}^{3}}} \\ & \Rightarrow y'=\dfrac{{{\left( 1-x \right)}^{2}}{{\left( 1+x \right)}^{3}}-x\left[ -2\left( 1-x \right){{\left( 1+x \right)}^{3}}+{{\left( 1-x \right)}^{2}}.3{{\left( 1+x \right)}^{2}} \right]}{{{\left( 1-x \right)}^{4}}{{\left( 1+x \right)}^{6}}} \\ & =\dfrac{{{\left( 1-x \right)}^{2}}{{\left( 1+x \right)}^{3}}-x\left( 1-{{x}^{2}} \right)\left[ -2\left( 1+2x+{{x}^{2}} \right)+3\left( 1-{{x}^{2}} \right) \right]}{{{\left( 1-x \right)}^{4}}{{\left( 1+x \right)}^{6}}} \\ & =\dfrac{{{\left( 1-x \right)}^{2}}{{\left( 1+x \right)}^{3}}-x\left( 1-{{x}^{2}} \right)\left( -5{{x}^{2}}-4x+1 \right)}{{{\left( 1-x \right)}^{4}}{{\left( 1+x \right)}^{6}}} \\ & =\dfrac{{{\left( 1-x \right)}^{2}}{{\left( 1+x \right)}^{3}}-x\left( 1-{{x}^{2}} \right)\left( 1+x \right)\left( 1-5x \right)}{{{\left( 1-x \right)}^{4}}{{\left( 1+x \right)}^{6}}} \\ & =\dfrac{\left( 1-x \right){{\left( 1+x \right)}^{2}}\left[ 1-{{x}^{2}}-x\left( 1-5x \right) \right]}{{{\left( 1-x \right)}^{4}}{{\left( 1+x \right)}^{6}}} \\ & =\dfrac{4{{x}^{2}}-x+1}{{{\left( 1-x \right)}^{3}}{{\left( 1+x \right)}^{4}}} \\ \end{aligned}\)
5.64
\(\begin{aligned} & y=\dfrac{\left( 2-{{x}^{2}} \right)\left( 3-{{x}^{3}} \right)}{{{\left( 1-x \right)}^{2}}}=\dfrac{{{x}^{5}}-2{{x}^{3}}-3{{x}^{2}}+6}{{{\left( 1-x \right)}^{2}}} \\ & \Rightarrow y'=\dfrac{\left( 5{{x}^{4}}-6{{x}^{2}}-6x \right){{\left( 1-x \right)}^{2}}+\left( {{x}^{5}}-2{{x}^{3}}-3{{x}^{2}}+6 \right).2.\left( 1-x \right)}{{{\left( 1-x \right)}^{4}}} \\ & =\dfrac{\left( 1-x \right)\left[ \left( 5{{x}^{4}}-6{{x}^{2}}-6x \right)\left( 1-x \right)+2{{x}^{5}}-4{{x}^{3}}-6{{x}^{2}}+12 \right]}{{{\left( 1-x \right)}^{4}}} \\ & =\dfrac{\left( 1-x \right)\left( -3{{x}^{5}}+5{{x}^{4}}+2{{x}^{3}}-6{{x}^{2}}-6x+12 \right)}{{{\left( 1-x \right)}^{4}}} \\ & =\dfrac{-3{{x}^{5}}+5{{x}^{4}}+2{{x}^{3}}-6{{x}^{2}}-6x+12}{{{\left( 1-x \right)}^{3}}} \\ \end{aligned} \)