\(A=\dfrac{2^4\cdot5^4\cdot3^4-2^4\cdot3^2\cdot5^2}{2^8\cdot3^3\cdot5^2}\)

\(=\dfrac{2^4\cdot3^2\cdot5^2\left(3^2\cdot5^2-1\right)}{2^8\cdot3^3\cdot5^2}=\dfrac{1}{16}\cdot\dfrac{1}{3}\cdot\dfrac{15^2-1}{1}\)

\(=\dfrac{224}{48}=\dfrac{14}{3}\)

Ta có: \(12^{10}\cdot35+2^{10}\cdot65+6^2\cdot12^6\cdot15^2\)

\(=2^{20}\cdot3^{10}\cdot5\cdot7+2^{10}\cdot5\cdot13+2^2\cdot3^2\cdot2^{12}\cdot3^6\cdot3^2\cdot5^2\)

\(=2^{20}\cdot3^{10}\cdot5\cdot7+2^{10}\cdot5\cdot13+2^{14}\cdot3^{10}\cdot5^2\)

\(=2^{10}\cdot5\left(3^{10}\cdot7+13+2^4\cdot5\right)\)

Ta có: \(\frac{12^{10}\cdot35+2^{10}\cdot65+6^2\cdot12^6\cdot15^2}{2^8\cdot10^4}\)

\(=\frac{2^{10}\cdot5\left(3^{10}\cdot7+13+2^4\cdot5\right)}{2^8\cdot2^4\cdot5^4}=\frac{2^{10}\cdot5\cdot413436}{2^{12}\cdot5^4}=\frac{413436}{2^2\cdot5^3}=\frac{413436}{500}=\frac{103359}{125}\)

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

\(\left\{{}\begin{matrix}\dfrac{2018}{2019}< 1\\\dfrac{2019}{2020}< 1\\\dfrac{2020}{2021}< 1\\\dfrac{2021}{2022}< 1\end{matrix}\right.\)

\(\Rightarrow\dfrac{2018}{2019}+\dfrac{2019}{2020}+\dfrac{2020}{2021}+\dfrac{2021}{2022}< 4\)

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 = \(\dfrac{2^9-3}{2^8}\)

Ta có: \(5^2\cdot6^{11}\cdot16^2+6^2\cdot12^6\cdot15^2\)

\(=5^2\cdot2^{11}\cdot3^{11}\cdot2^8+2^2\cdot3^2\cdot2^{12}\cdot3^6\cdot5^2\cdot3^2\)

\(=2^{19}\cdot3^{11}\cdot5^2+2^{14}\cdot3^{10}\cdot5^2\)

\(=2^{14}\cdot3^{10}\cdot5^2\left(2^5\cdot3+1\right)=2^{14}\cdot3^{10}\cdot5^2\cdot97\)

Ta có: \(2\cdot6^{12}\cdot10^4-81^2\cdot960^3\)

\(=2\cdot2^{12}\cdot3^{12}\cdot2^4\cdot5^4-3^8\cdot\left(2^6\cdot3\cdot5\right)^3\)

\(=2^{17}\cdot3^{12}\cdot5^4-2^{18}\cdot3^{11}\cdot5^3\)

\(=2^{17}\cdot3^{11}\cdot5^3\left(3\cdot5-2\right)=2^{17}\cdot3^{11}\cdot5^3\cdot13\)

Ta có: \(\frac{5^2\cdot6^{11}\cdot16^2+6^2\cdot12^6\cdot15^2}{2\cdot6^{12}\cdot10^4-81^2\cdot960^3}\)

\(=\frac{2^{14}\cdot3^{10}\cdot5^2\cdot97}{2^{17}\cdot3^{11}\cdot5^3\cdot13}=\frac{97}{13\cdot2^3\cdot3\cdot5}=\frac{97}{1560}\)

dấu ; nghĩa là gì vậy bạn

4/16-5/10+2

=1/4+1/2+2

= 3/4+2

=11/4

mình làm chi tiết nha bn

\(\dfrac{4}{16}-\dfrac{5}{10}+2=\dfrac{20}{80}-\dfrac{40}{80}+\dfrac{160}{80}=\dfrac{140}{80}=\dfrac{7}{4}\)

`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`