tính giá trị biểu thức B=x^10+20^9+20^8+…+20^2+20x+20 với x=-19 14/08/2021 Bởi Amaya tính giá trị biểu thức B=x^10+20^9+20^8+…+20^2+20x+20 với x=-19
Đáp án: $\begin{array}{l}B = {x^{10}} + {20^9} + {20^8} + … + {20^2} + 20x + 20\\ = {x^{10}} + 20x + \left( {{{20}^9} + {{20}^8} + … + {{20}^2} + 20} \right)\\ = {x^{10}} + 20x + A\\A = {20^9} + {20^8} + .. + {20^2} + 20\\ \Rightarrow 20A = {20^{10}} + {20^9} + … + {20^3} + {20^2}\\ \Rightarrow 20A – A = {20^{10}} – 20\\ \Rightarrow 19A = {20^{10}} – 20\\ \Rightarrow A = \dfrac{{{{20}^{10}} – 20}}{{19}}\\ \Rightarrow B = {x^{10}} + 20x + \dfrac{{{{20}^{10}} – 20}}{{19}}\\x = – 19\\ \Rightarrow B = {\left( { – 19} \right)^{10}} + 20.\left( { – 19} \right) + \dfrac{{{{20}^{10}} – 20}}{{19}}\\ = {19^{10}} + \dfrac{{{{20}^{10}} – 20}}{{19}} – 380\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
B = {x^{10}} + {20^9} + {20^8} + … + {20^2} + 20x + 20\\
= {x^{10}} + 20x + \left( {{{20}^9} + {{20}^8} + … + {{20}^2} + 20} \right)\\
= {x^{10}} + 20x + A\\
A = {20^9} + {20^8} + .. + {20^2} + 20\\
\Rightarrow 20A = {20^{10}} + {20^9} + … + {20^3} + {20^2}\\
\Rightarrow 20A – A = {20^{10}} – 20\\
\Rightarrow 19A = {20^{10}} – 20\\
\Rightarrow A = \dfrac{{{{20}^{10}} – 20}}{{19}}\\
\Rightarrow B = {x^{10}} + 20x + \dfrac{{{{20}^{10}} – 20}}{{19}}\\
x = – 19\\
\Rightarrow B = {\left( { – 19} \right)^{10}} + 20.\left( { – 19} \right) + \dfrac{{{{20}^{10}} – 20}}{{19}}\\
= {19^{10}} + \dfrac{{{{20}^{10}} – 20}}{{19}} – 380
\end{array}$