`Answer:`

\(A=1\frac{3}{5}+\left(3\frac{2}{5}-2\right)=\frac{8}{5}+\frac{17}{5}-2=\frac{25}{5}-2=3\)

\(B=\frac{3}{7}+\left(-\frac{1}{5}+-\frac{3}{7}\right)=\frac{3}{7}-\frac{1}{5}-\frac{3}{7}=\left(\frac{3}{7}-\frac{3}{7}\right)-\frac{1}{5}=-\frac{1}{5}\)

\(C=\left(6-2\frac{4}{5}\right)3\frac{1}{8}-1\frac{3}{5}:\frac{1}{4}=\frac{16}{5}.\frac{25}{8}-\frac{32}{5}=10-\frac{32}{5}=\frac{18}{5}\)

\(D=\left(-\frac{5}{24}+0,75+\frac{7}{12}\right):\left(-2\frac{1}{8}\right)=\left(-\frac{5}{24}+\frac{18}{24}+\frac{14}{24}\right):-\frac{1}{4}=\frac{9}{8}.-\frac{4}{1}=-\frac{9}{2}\)

\(E=-\frac{5}{7}.\frac{2}{11}+-\frac{5}{7}.\frac{9}{11}+1\frac{5}{7}=-\frac{5}{7}.\left(\frac{2}{11}+\frac{9}{11}\right)+\frac{12}{7}=-\frac{5}{7}+\frac{12}{7}=1\)

\(F=\frac{6}{7}+\frac{5}{8}:5-\frac{3}{16}.\left(-2\right)^2=\frac{6}{7}+\frac{5}{40}-\frac{12}{16}=\frac{48}{56}+\frac{7}{56}-\frac{42}{56}=\frac{31}{28}\)

<=> (7/2 + 2x) . 8/3 = 16/3

<=>  28/3 + 16/3x = 16/3 

<=> 16/3x = 4

<=> x = 0,75 

\(4^x+4^{x+3}=4160\)

\(4^x+4^x.4^3=4160\)

\(4^x\left(1+4^3\right)=4160\)

\(4^x=\dfrac{4160}{65}=64=4^3\)

\(\rightarrow x=3\)

`Answer:`

Ta thấy: 

\(9=1.9\)

\(20=10.2\)

\(33=11.3\)

...

\(9200=100.92\)

`=>` Mẫu thức của từng nhân tử có dạng là \(n\left(n+8\right)\)

Xét dạng tổng quát của nhân tử: \(1+\frac{7}{n\left(n+8\right)}=\frac{n^2+8n+7}{n\left(n+8\right)}=\frac{\left(n+1\right)\left(n+7\right)}{n\left(n+8\right)}\)

\(n=1\Rightarrow1+\frac{7}{1.9}=\frac{2.8}{1.9}\)

\(n=2\Rightarrow1+\frac{7}{2.10}=\frac{3.9}{2.10}\)

\(n=3\Rightarrow1=\frac{7}{3.10}=\frac{4.10}{3.11}\)

...

\(n=92\Rightarrow1+\frac{7}{92.100}=\frac{93.99}{92.100}\)

\(\Rightarrow\frac{2.8}{1.9}.\frac{3.9}{2.10}.\frac{4.10}{3.11}...\frac{93.99}{92.100}=\frac{\left(2.3.4...93\right)\left(8.9.10...9\right)}{\left(1.2.3...92\right)\left(9.10.11...100\right)}=\frac{93.8}{1.100}=\frac{186}{25}\)

Mọi người đánh giúp mình nhé! Hạn là tối nay!!