cho A=1+1/2+(1/2)^2+(1/2)^3+(1/2)^4+(1/2)^5.a) tính 2A. b) Chứng minh A=-(1/2)+2 30/11/2021 Bởi Eden cho A=1+1/2+(1/2)^2+(1/2)^3+(1/2)^4+(1/2)^5.a) tính 2A. b) Chứng minh A=-(1/2)+2
Đáp án: $\begin{array}{l}a)A = 1 + \dfrac{1}{2} + {\left( {\dfrac{1}{2}} \right)^2} + {\left( {\dfrac{1}{2}} \right)^3} + {\left( {\dfrac{1}{2}} \right)^4} + {\left( {\dfrac{1}{2}} \right)^5}\\ \Rightarrow 2.A = 2 + 1 + {\left( {\dfrac{1}{2}} \right)^2} + {\left( {\dfrac{1}{2}} \right)^3} + {\left( {\dfrac{1}{2}} \right)^4}\\b)2A – A\\ = \left( {2 + 1 + {{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{1}{2}} \right)}^3} + {{\left( {\dfrac{1}{2}} \right)}^4}} \right)\\ – \left( {1 + \dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{1}{2}} \right)}^3} + {{\left( {\dfrac{1}{2}} \right)}^4} + \left( {\dfrac{1}{2}} \right)} \right)\\ \Rightarrow A = 2 – {\left( {\dfrac{1}{2}} \right)^5}\\ = – {\left( {\dfrac{1}{2}} \right)^5} + 2\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
a)A = 1 + \dfrac{1}{2} + {\left( {\dfrac{1}{2}} \right)^2} + {\left( {\dfrac{1}{2}} \right)^3} + {\left( {\dfrac{1}{2}} \right)^4} + {\left( {\dfrac{1}{2}} \right)^5}\\
\Rightarrow 2.A = 2 + 1 + {\left( {\dfrac{1}{2}} \right)^2} + {\left( {\dfrac{1}{2}} \right)^3} + {\left( {\dfrac{1}{2}} \right)^4}\\
b)2A – A\\
= \left( {2 + 1 + {{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{1}{2}} \right)}^3} + {{\left( {\dfrac{1}{2}} \right)}^4}} \right)\\
– \left( {1 + \dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{1}{2}} \right)}^3} + {{\left( {\dfrac{1}{2}} \right)}^4} + \left( {\dfrac{1}{2}} \right)} \right)\\
\Rightarrow A = 2 – {\left( {\dfrac{1}{2}} \right)^5}\\
= – {\left( {\dfrac{1}{2}} \right)^5} + 2
\end{array}$