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\(B=1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{200}{2^{200}}\)
\(2B=2\left(1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{200}{2^{200}}\right)\)
\(2B=2+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{200}{2^{199}}\)
\(2B-B=\left(2+\frac{3}{2^2}+...+\frac{200}{2^{199}}\right)-\left(1+\frac{3}{2^3}+...+\frac{200}{2^{200}}\right)\)
.... đặt A=... giiả tiếp
a.
\(\frac{2^7\times9^3}{6^5\times8^2}=\frac{2^7\times\left(3^2\right)^3}{\left(2\times3\right)^5\times\left(2^3\right)^2}=\frac{2^7\times3^6}{2^5\times3^5\times2^6}=\frac{3}{2^4}=\frac{3}{16}\)
b.
\(\frac{6^3+3\times6^2+3^3}{-13}=\frac{\left(2\times3\right)^3+3\times\left(3\times2\right)^2+3^3}{-13}=\frac{2^3\times3^3+3\times3^2\times2^2+3^3}{-13}=\frac{8\times3^3+3^3\times4+3^3}{-13}\)\(=\frac{3^3\times\left(8+4+1\right)}{-13}=\frac{27\times13}{-13}=-27\)
c.
\(\frac{5^4\times20^4}{25^5\times4^5}=\frac{\left(5\times20\right)^4}{\left(25\times4\right)^5}=\frac{100^4}{100^5}=\frac{1}{100}\)
d.
\(\left(\frac{5^4-5^3}{125^4}\right)=\frac{5^3\times\left(5-1\right)}{\left(5^3\right)^4}=\frac{5^3\times4}{5^{12}}=\frac{4}{5^9}\)
a)\(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.\left(3^2\right)^3}{2^5.3^5.\left(2^3\right)^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{3}{2^4}\)
b)\(\frac{6^3+3.6^2+3^3}{-13}=\frac{6.6^2+3.6^2+3^3}{-13}=\frac{6^2.\left(6+3\right)+3^3}{-13}=\frac{6^2.9+3^3}{-13}=\frac{6^2.3^2+3.3^2}{-13}=\frac{3^2.\left(6^2+3\right)}{-13}=\frac{3^2.39}{-13}=3^2.\left(-3\right)=-27\)
c)\(\frac{5^4.20^4}{25^5.4^5}=\frac{100^4}{100^5}=\frac{1}{100}\)
a, Dat A =\(\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-...-\frac{1}{3^{198}}+\frac{1}{3^{199}}\)
\(\Rightarrow\frac{1}{3}A=\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...-\frac{1}{3^{199}}+\frac{1}{3^{200}}\)
\(\Rightarrow\frac{1}{3}A+A=\left(\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...-\frac{1}{3^{199}}+\frac{1}{3^{200}}\right)+\left(\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-...-\frac{1}{3^{198}}+\frac{1}{3^{199}}\right)\)
\(\Rightarrow\frac{4}{3}A=\frac{1}{3}+\frac{1}{3^{200}}\)
\(\Rightarrow A=\frac{\frac{1}{3}+\frac{1}{3^{200}}}{\frac{4}{3}}\)
chung minh tuong tu cau b va c
a, \(\dfrac{2\cdot8^4\cdot27^2+4\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)
=\(\dfrac{2\cdot\left(2^3\right)^4\cdot\left(3^3\right)^2+2^2\cdot2^9\cdot3^9}{2^7\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot\left(3^2\right)^4}\)
=\(\dfrac{2\cdot2^{12}\cdot3^6+2^{11}\cdot3^9}{2^{14}\cdot3^7+2^{10}\cdot5\cdot3^8}\)
=\(\dfrac{2^{11}\cdot3^6\cdot\left(2^2+3^3\right)}{2^{10}\cdot3^7\cdot\left(2^4+5\cdot3\right)}\)
=\(\dfrac{2^{11}\cdot3^6\cdot31}{2^{10}\cdot3^7\cdot31}\)
=\(\dfrac{2}{3}\)
b, \(\dfrac{\dfrac{8}{27}\cdot\dfrac{9}{16}\cdot\left(-1\right)}{\dfrac{4}{25}\cdot\dfrac{-125}{1728}}\)
=\(\dfrac{\dfrac{8\cdot9\cdot\left(-1\right)}{27\cdot16}}{\dfrac{4\cdot\left(-125\right)}{25\cdot1728}}\)
=\(\dfrac{\dfrac{-1}{6}}{\dfrac{-5}{432}}\)
=\(\dfrac{-1}{6}\cdot\dfrac{-432}{5}\)
=\(\dfrac{72}{5}\)
a) Ta có: A = 1 + 3 + 32 + ... + 3200
⇒ 3A = 3 + 32 + 33 + ... + 3200
⇒ 3A - A = (3 + 32 + 33 + ... + 3201) - (1 + 3 + 32 + ... + 3200)
⇒ 2A = 3201 - 1
⇒ A = 2A : A = \(\dfrac{3^{201}-1}{2}\)
Vậy A = \(\dfrac{3^{201}-1}{3}\)
Khác nhau quá trời, trên kia 3 dưới 2 là sao bạn???