cho A=3 mũ 1+3 mũ 2+3 mũ 3+…….+3 mũ 2010 a)thu gọn A b)Tìm x để 2A+3=3 mũ x 15/08/2021 Bởi Melanie cho A=3 mũ 1+3 mũ 2+3 mũ 3+…….+3 mũ 2010 a)thu gọn A b)Tìm x để 2A+3=3 mũ x
Đáp án: \(\begin{array}{l}a)\,\,\,A = \frac{{{3^{2011}} – 3}}{2}\\b)\,\,x = 2011\end{array}\) Giải thích các bước giải: \(A = {3^1} + {3^2} + {3^3} + …. + {3^{2010}}\) a) Thu gọn A: Ta có: \(3A = 3\left( {{3^1} + {3^2} + {3^3} + …… + {3^{2010}}} \right) = {3^2} + {3^3} + {3^4} + …. + {3^{2010}}\) \(\begin{array}{l} \Rightarrow 3A – A = {3^2} + {3^3} + {3^4} + …. + {3^{2011}} – \left( {{3^1} + {3^2} + {3^3} + …. + {3^{2010}}} \right)\\ \Rightarrow 2A = {3^{2011}} – {3^1} = {3^{2011}} – 3\\ \Rightarrow A = \frac{{{3^{2011}} – 3}}{2}.\end{array}\) b) Ta có: \(\begin{array}{l}2A + 3 = {3^x}\\ \Leftrightarrow 2.\frac{{{3^{2011}} – 3}}{2} + 3 = {3^x}\\ \Leftrightarrow {3^{2011}} – 3 + 3 = {3^x}\\ \Leftrightarrow {3^{2011}} = {3^x}\\ \Leftrightarrow x = 2011.\end{array}\) Bình luận
a) `A=3+3^2+…+3^2010` `3A=3^2+3^3+…+3^2011` `3A-A=3^2+3^3+…+3^2011-(3+3^2+…+3^2010)` `2A=3^2011-3` `A=(3^2011-3)/2` b) `2A+3=3^x` `⇔2. (3^2011-3)/2=3^x` `⇔3^2011-3+3=3^x` `⇔3^2011=3^x` `⇔x=2011` Bình luận
Đáp án:
\(\begin{array}{l}
a)\,\,\,A = \frac{{{3^{2011}} – 3}}{2}\\
b)\,\,x = 2011
\end{array}\)
Giải thích các bước giải:
\(A = {3^1} + {3^2} + {3^3} + …. + {3^{2010}}\)
a) Thu gọn A:
Ta có: \(3A = 3\left( {{3^1} + {3^2} + {3^3} + …… + {3^{2010}}} \right) = {3^2} + {3^3} + {3^4} + …. + {3^{2010}}\)
\(\begin{array}{l} \Rightarrow 3A – A = {3^2} + {3^3} + {3^4} + …. + {3^{2011}} – \left( {{3^1} + {3^2} + {3^3} + …. + {3^{2010}}} \right)\\ \Rightarrow 2A = {3^{2011}} – {3^1} = {3^{2011}} – 3\\ \Rightarrow A = \frac{{{3^{2011}} – 3}}{2}.\end{array}\)
b) Ta có:
\(\begin{array}{l}2A + 3 = {3^x}\\ \Leftrightarrow 2.\frac{{{3^{2011}} – 3}}{2} + 3 = {3^x}\\ \Leftrightarrow {3^{2011}} – 3 + 3 = {3^x}\\ \Leftrightarrow {3^{2011}} = {3^x}\\ \Leftrightarrow x = 2011.\end{array}\)
a)
`A=3+3^2+…+3^2010`
`3A=3^2+3^3+…+3^2011`
`3A-A=3^2+3^3+…+3^2011-(3+3^2+…+3^2010)`
`2A=3^2011-3`
`A=(3^2011-3)/2`
b)
`2A+3=3^x`
`⇔2. (3^2011-3)/2=3^x`
`⇔3^2011-3+3=3^x`
`⇔3^2011=3^x`
`⇔x=2011`