\(\dfrac{10^4\cdot81-16\cdot15^2}{4^4\cdot675}\).Tính
rut gon A=\(\dfrac{10^4\cdot81-16\cdot15^2}{4^4\cdot675}\)
\(A=\dfrac{2^4\cdot5^4\cdot3^4-2^4\cdot3^2\cdot5^2}{2^8\cdot3^3\cdot5^2}\)
\(=\dfrac{2^4\cdot3^2\cdot5^2\left(3^2\cdot5^2-1\right)}{2^8\cdot3^3\cdot5^2}=\dfrac{1}{16}\cdot\dfrac{1}{3}\cdot\dfrac{15^2-1}{1}\)
\(=\dfrac{224}{48}=\dfrac{14}{3}\)
tính :
\(\dfrac{12^{10}\cdot35+2^{10}\cdot65+6^2\cdot12^6\cdot15^2}{2^8\cdot10^4}\)
a) Tìm số tự nhiên n biết:
\(\dfrac{4}{3\cdot5}+\dfrac{8}{5\cdot9}+\dfrac{12}{9\cdot15}+....+\dfrac{32}{n\cdot\left(n+16\right)}=\dfrac{16}{25}\)
b) Chứng tỏ rằng:
\(\dfrac{2018}{2019}+\dfrac{2019}{2020}+\dfrac{2020}{2021}+\dfrac{2021}{2018}>4\)
a) \(2\left(\dfrac{2}{3.5}+\dfrac{4}{5.9}+...+\dfrac{16}{n\left(n+16\right)}\right)=\dfrac{16}{25}\)
\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{n}-\dfrac{1}{n+16}=\dfrac{8}{25}\)
\(\dfrac{1}{3}-\dfrac{1}{n+16}=\dfrac{8}{25}\)
\(\dfrac{n+13}{3\left(n+16\right)}=\dfrac{8}{25}\)
\(24n+384=25n+325\)
\(25n-24n=384-325\)
\(n=59\)
b) Sai đề nha
\(\left\{{}\begin{matrix}\dfrac{2018}{2019}< 1\\\dfrac{2019}{2020}< 1\\\dfrac{2020}{2021}< 1\\\dfrac{2021}{2022}< 1\end{matrix}\right.\)
\(\Rightarrow\dfrac{2018}{2019}+\dfrac{2019}{2020}+\dfrac{2020}{2021}+\dfrac{2021}{2022}< 4\)
chị ơi hình như chị nhầm rồi P/s cuối phải là 1/n.(n+6)thì phải
Tính tổng S = \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{8}\) + \(\dfrac{4}{16}\) + ... + \(\dfrac{10}{2^{10}}\)
Ta có S = \(\dfrac{1}{2}+\dfrac{2}{4}+\dfrac{3}{8}+\dfrac{4}{16}+...+\dfrac{10}{2^{10}}\)
= \(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+\dfrac{4}{2^4}+...+\dfrac{10}{2^{10}}\)
2S = 1 + \(\dfrac{2}{2}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{10}{2^9}\)
2S - S = ( 1 + \(\dfrac{2}{2}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{10}{2^9}\)) - ( \(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+\dfrac{4}{2^4}+...+\dfrac{10}{2^{10}}\))
S = 1 + \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}-\dfrac{10}{2^{10}}\)
Đặt A = 1 + \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}\)
2A = 2 + 1 + \(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^8}\)
2A - A = ( 2 + 1 + \(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^8}\)) - ( 1 + \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}\))
A = 2 - \(\dfrac{1}{2^9}\)
⇒ S = 2 - \(\dfrac{1}{2^9}\) - \(\dfrac{10}{2^{10}}\) = \(\dfrac{2^{11}}{2^{10}}-\dfrac{2}{2^{10}}-\dfrac{10}{2^{10}}=\dfrac{2^2\left(2^9-3\right)}{2^{10}}=\dfrac{2^9-3}{2^8}\)
Vậy S = \(\dfrac{2^9-3}{2^8}\)
tính :
\(\dfrac{5^2\cdot6^{11}\cdot16^2+6^2\cdot12^6\cdot15^2}{2\cdot6^{12}\cdot10^4-81^2\cdot960^3}\)
tính tổng S = \(\dfrac{1}{2}+\dfrac{2}{4}+\dfrac{3}{8}+\dfrac{4}{16}+...+\dfrac{10}{2^{10}}\)
làm nhanh hộ mình với
\(A=\frac{2^2\cdot3\cdot2^{16}\cdot81}{3^2\cdot2^{15}\cdot4+4^9;27}\)
Tính :
\(\dfrac{4}{16}-\dfrac{5}{10}+2\)
mình làm chi tiết nha bn
\(\dfrac{4}{16}-\dfrac{5}{10}+2=\dfrac{20}{80}-\dfrac{40}{80}+\dfrac{160}{80}=\dfrac{140}{80}=\dfrac{7}{4}\)
\(\dfrac{4}{16}-\dfrac{5}{10}+2\)
=\(\dfrac{1}{4}+\dfrac{1}{2}+2\)
=\(\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{8}{4}\)
=\(\dfrac{11}{4}\)
thực hiện phép tính
\(\dfrac{16}{103}\)+(38+\(\dfrac{-16}{103}\))
\(\dfrac{100}{91}\)+310+\(\dfrac{-9}{91}\)
\(\dfrac{-13}{49}\)+(\(\dfrac{-36}{49}\)+41)
\(\dfrac{-10}{71}\):1\(\dfrac{4}{71}\)
\(\dfrac{4}{15}\)+\(\dfrac{8}{15}\):2 -\(\dfrac{1}{18}\).\(\left(-3\right)^2\)
`16/803+38+(-16/803)`
`=16/803-16/803+38`
`=0+38=38`
`100/91+310-9/91`
`=100/91-9/91+310`
`=1+310=311`
\(\dfrac{16}{103}+\left(38+\dfrac{-16}{103}\right)\)
\(=\dfrac{16}{103}+38+\dfrac{-16}{103}\)
\(=\dfrac{16}{103}+\dfrac{-16}{103}+38\)
\(=0+38\)
\(=38\)