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Câu hỏi: Giá trị của $\underset{x\to -1}{\mathop{\lim }} \dfrac{{{\left( {{x}^{2}}-3x+2 \right)}^{2019}}+{{\left( 2x+8 \right)}^{2019}}-{{2.6}^{2019}}}{\left( x+1 \right)\left( 2020+x \right)}$ bằng
A. $-\dfrac{{{3.6}^{2019}}}{2019}.$
B. ${{5.6}^{2018}}.$
C. $-{{3.6}^{2018}}.$
D. $\dfrac{{{7.6}^{2018}}}{2019}.$
A. $-\dfrac{{{3.6}^{2019}}}{2019}.$
B. ${{5.6}^{2018}}.$
C. $-{{3.6}^{2018}}.$
D. $\dfrac{{{7.6}^{2018}}}{2019}.$
Xét hàm số $h\left( x \right)={{\left( {{x}^{2}}-3x+2 \right)}^{2019}}+{{\left( 2x+8 \right)}^{2019}}$ liên tục trên $\mathbb{R}$ thì $h\left( -1 \right)={{2.6}^{2019}}$
Theo định nghĩa đạo hàm của hàm số ta có ${h}'\left( {{x}_{0}} \right)=\underset{x\to {{x}_{0}}}{\mathop{\lim }} \dfrac{h\left( x \right)-h\left( {{x}_{0}} \right)}{x-{{x}_{0}}}$
Do đó $\underset{x\to -1}{\mathop{\lim }} \dfrac{{{\left( {{x}^{2}}-3x+2 \right)}^{2019}}+{{\left( 2x+8 \right)}^{2019}}-{{2.6}^{2019}}}{\left( x+1 \right)\left( 2020+x \right)}=\underset{x\to -1}{\mathop{\lim }} \dfrac{1}{2019}.\dfrac{h\left( x \right)-h\left( -1 \right)}{x-\left( -1 \right)}=\dfrac{1}{2019}{h}'\left( -1 \right)$
Ta có ${h}'\left( x \right)=2019\left( 2x-3 \right){{\left( {{x}^{2}}-3x+2 \right)}^{2018}}+2019.2.{{\left( 2x+8 \right)}^{2018}}\Rightarrow {h}'\left( -1 \right)=-{{6057.6}^{2018}}$
Vậy ta có $\underset{x\to -1}{\mathop{\lim }} \dfrac{{{\left( {{x}^{2}}-3x+2 \right)}^{2019}}+{{\left( 2x+8 \right)}^{2019}}-{{2.6}^{2019}}}{\left( x+1 \right)\left( 2020+x \right)}=\dfrac{1}{2019}{h}'\left( -1 \right)=-{{3.6}^{2018}}.$
Theo định nghĩa đạo hàm của hàm số ta có ${h}'\left( {{x}_{0}} \right)=\underset{x\to {{x}_{0}}}{\mathop{\lim }} \dfrac{h\left( x \right)-h\left( {{x}_{0}} \right)}{x-{{x}_{0}}}$
Do đó $\underset{x\to -1}{\mathop{\lim }} \dfrac{{{\left( {{x}^{2}}-3x+2 \right)}^{2019}}+{{\left( 2x+8 \right)}^{2019}}-{{2.6}^{2019}}}{\left( x+1 \right)\left( 2020+x \right)}=\underset{x\to -1}{\mathop{\lim }} \dfrac{1}{2019}.\dfrac{h\left( x \right)-h\left( -1 \right)}{x-\left( -1 \right)}=\dfrac{1}{2019}{h}'\left( -1 \right)$
Ta có ${h}'\left( x \right)=2019\left( 2x-3 \right){{\left( {{x}^{2}}-3x+2 \right)}^{2018}}+2019.2.{{\left( 2x+8 \right)}^{2018}}\Rightarrow {h}'\left( -1 \right)=-{{6057.6}^{2018}}$
Vậy ta có $\underset{x\to -1}{\mathop{\lim }} \dfrac{{{\left( {{x}^{2}}-3x+2 \right)}^{2019}}+{{\left( 2x+8 \right)}^{2019}}-{{2.6}^{2019}}}{\left( x+1 \right)\left( 2020+x \right)}=\dfrac{1}{2019}{h}'\left( -1 \right)=-{{3.6}^{2018}}.$
Đáp án C.