\(PTK_{Cu\left(OH\right)_2}=64+\left(16+1\right).2=98\left(đvC\right)\)

\(\%m_{Cu}=\dfrac{64}{98}.100=65,30\%\)

\(\%m_O=\dfrac{16.2}{98}.100=32,65\%\)

\(\%m_H=\dfrac{1.2}{98}.100=2,04\%\)

\(PTK_{H_2SO_4}=2.1+32+4.16=98\left(đvC\right)\)

\(\%m_H=\dfrac{2.1}{98}.100=2,04\%\)

\(\%m_S=\dfrac{32}{98}.100=32,65\%\)

\(\%m_O=\dfrac{4.16}{98}.100=65,30\%\)

các ý còn lại làm tương tự

*Giống:sống theo đàn

*Khác:

-Khỉ:có chai mông lớn,túi má lớn,đuôi dài

-Vượn:Có chai mông nhỏ,khong có túi má và đuôi

1  A

2 D

3B

4 C

5 B

Em cần mọi người hỗ trợ những câu nào hay toàn đề em nhỉ? Hay em đăng lên cho các bạn tham khảo đề nè!

Xét: \(\frac{\left(17^{2017}+16^{2017}\right)^{2018}}{17^{2017.2018}}=\left(\frac{17^{2017}+16^{2017}}{17^{2017}}\right)^{2018}=\left(1+\left(\frac{16}{17}\right)^{2017}\right)^{2018}\)

\(\frac{\left(17^{2018}+16^{2018}\right)^{2017}}{17^{2017.2018}}=\left(\frac{17^{2018}+16^{2018}}{17^{2018}}\right)^{2017}=\left(1+\left(\frac{16}{17}\right)^{2018}\right)^{2017}\)

Ta có: \(0< \frac{16}{17}< 1\)

=> \(\left(\frac{16}{17}\right)^{2017}>\left(\frac{16}{17}\right)^{2018}\)

=> \(1+\left(\frac{16}{17}\right)^{2017}>1+\left(\frac{16}{17}\right)^{2018}>1\)

=> \(\left(1+\left(\frac{16}{17}\right)^{2017}\right)^{2018}>\left(1+\left(\frac{16}{17}\right)^{2018}\right)^{2017}\)

=> \(\left(17^{2017}+16^{2017}\right)^{2018}>\left(17^{2018}+16^{2018}\right)^{2017}\)

a: \(\left(\frac{1}{24}-\frac{3}{16}\right):\left(-\frac38+\frac12\right)\)

\(=\left(\frac{2}{48}-\frac{9}{48}\right):\left(-\frac38+\frac48\right)\)

\(=\frac{-7}{48}:\frac18=-\frac{7}{48}\cdot8=-\frac76\)

b: \(-\frac57:\left(5-4\frac47\right)+\left(15\frac13\cdot\frac{-1}{23}\right)\)

\(=-\frac57:\left(5-4-\frac47\right)+\frac{46}{3}\cdot\frac{-1}{23}=-\frac57:\left(1-\frac47\right)+\frac{-2}{3}\)

\(=-\frac57:\frac37+\frac{-2}{3}=-\frac53-\frac23=-\frac73\)

c: \(\left(-2\frac23\cdot1\frac45+2\frac56\right):\left(\frac16-\frac34\right)\)

\(=\left(-\frac83\cdot\frac95+\frac{17}{6}\right):\left(\frac{2}{12}-\frac{9}{12}\right)\)

\(=\left(-\frac{24}{5}+\frac{17}{6}\right):\frac{-7}{12}=\left(-\frac{144}{30}+\frac{85}{30}\right):\frac{-7}{12}\)

\(=-\frac{59}{30}\cdot\frac{12}{-7}=\frac{59\cdot2}{7\cdot5}=\frac{118}{35}\)

d: \(\left(\frac{1}{3^2}-\frac{1}{7^2}\right)\left(\frac{1}{4^2}-\frac{1}{7^2}\right)\cdot\ldots\cdot\left(\frac{1}{49^2}-\frac{1}{7^2}\right)\)

\(=\left(\frac{1}{3^2}-\frac{1}{7^2}\right)\left(\frac{1}{4^2}-\frac{1}{7^2}\right)\cdot\ldots\cdot\left(\frac{1}{7^2}-\frac{1}{7^2}\right)\cdot\left(\frac{1}{8^2}-\frac{1}{7^2}\right)\cdot\ldots\cdot\left(\frac{1}{49^2}-\frac{1}{7^2}\right)\)

\(=\left(\frac{1}{3^2}-\frac{1}{7^2}\right)\left(\frac{1}{4^2}-\frac{1}{7^2}\right)\cdot\ldots\cdot\left(\frac{1}{49}-\frac{1}{49}\right)\cdot\left(\frac{1}{8^2}-\frac{1}{7^2}\right)\cdot\ldots\cdot\left(\frac{1}{49^2}-\frac{1}{7^2}\right)\)

\(=\left(\frac{1}{3^2}-\frac{1}{7^2}\right)\left(\frac{1}{4^2}-\frac{1}{7^2}\right)\cdot\ldots\cdot0\cdot\left(\frac{1}{8^2}-\frac{1}{7^2}\right)\cdot\ldots\cdot\left(\frac{1}{49^2}-\frac{1}{7^2}\right)\)

=0

e: \(\left(1-\frac23\right)\left(1-\frac25\right)\left(1-\frac27\right)\cdot\ldots\cdot\left(1-\frac{2}{2023}\right)\)

\(=\frac13\cdot\frac35\cdot\frac57\cdot\ldots\cdot\frac{2021}{2023}=\frac{1}{2023}\)

f: \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\cdot\ldots\cdot\left(\frac{1}{100^2}-1\right)\)

\(=\left(\frac12-1\right)\left(\frac13-1\right)\cdot\ldots\cdot\left(\frac{1}{100}-1\right)\left(\frac12+1\right)\left(\frac13+1\right)\cdot\ldots\cdot\left(\frac{1}{100}+1\right)\)

\(=\frac{-1}{2}\cdot\frac{-2}{3}\cdot\ldots\cdot\frac{-99}{100}\cdot\frac32\cdot\frac43\cdot\ldots\cdot\frac{101}{100}=\frac{-1}{100}\cdot\frac{101}{2}=-\frac{101}{200}\)

g:

Ta có công thức sau:

\(1-\frac{2}{n\left(n+1\right)}\)

\(=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}\)

\(=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)

\(\left(1-\frac13\right)\left(1-\frac16\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)\cdot\ldots\cdot\left(1-\frac{1}{780}\right)\)

\(=\left(1-\frac26\right)\left(1-\frac{2}{12}\right)\left(1-\frac{2}{20}\right)\left(1-\frac{2}{30}\right)\cdot\ldots\cdot\left(1-\frac{2}{1560}\right)\)

\(=\left(1-\frac{2}{2\cdot3}\right)\left(1-\frac{2}{3\cdot4}\right)\cdot\ldots\cdot\left(1-\frac{2}{39\cdot40}\right)\)

\(=\frac{\left(2+2\right)\left(2-1\right)}{2\left(2+1\right)}\cdot\frac{\left(3+2\right)\left(3-1\right)}{3\left(3+1\right)}\cdot\ldots\cdot\frac{\left(39+2\right)\left(39-1\right)}{39\left(39+1\right)}\)

\(=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot\ldots\cdot\frac{41\cdot38}{39\cdot40}=\frac{4\cdot5\cdot\ldots\cdot41}{3\cdot4\cdot..\cdot40}\cdot\frac{1\cdot2\cdot\ldots\cdot38}{2\cdot3\cdot\ldots\cdot39}\)

\(=\frac{41}{3}\cdot\frac{1}{39}=\frac{41}{117}\)