Thấy : \(\dfrac{1}{b^2+1}=1-\dfrac{b^2}{b^2+1}\ge1-\dfrac{b^2}{2b}=1-\dfrac{b}{2}\)  ( AD BĐT Cô-si ) 

Suy ra : \(\dfrac{a+1}{b^2+1}\ge\left(a+1\right)\left(1-\dfrac{b}{2}\right)=a+1-\dfrac{ab}{2}-\dfrac{b}{2}\)

CMTT : \(\dfrac{b+1}{c^2+1}\ge b+1-\dfrac{bc}{2}-\dfrac{c}{2}\)  ; \(\dfrac{c+1}{a^2+1}\ge c+1-\dfrac{ac}{2}-\dfrac{a}{2}\)

Suy ra : \(VT\ge\dfrac{a+b+c}{2}+3-\dfrac{ab+bc+ac}{2}=\dfrac{9}{2}-\dfrac{ab+bc+ac}{2}\)  (1)

Mặt khác : \(ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}=3\) (2)

(1) ; (2)  Suy ra :  \(VT\ge\dfrac{9}{2}-\dfrac{3}{2}=3=a+b+c\)

" = " \(\Leftrightarrow a=b=c=1\)

\(\dfrac{d_A}{H_2}< 17\Rightarrow M_A< 17.2=34\)

\(M_A=M_{\left(CH_3\right)_n}=\left(12+3\right).n=15n\)

\(A=1+3+3^2+...+3^{2020}\) . A có : 2021 số 

Ta có : \(A=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{2018}\left(1+3+3^2\right)\)

\(=13+13.3^3+...+13.3^{2018}\) 

\(=13\left(1+3^3+...+3^{2018}\right)⋮13\left(đpcm\right)\)

Với n = 2;3;...;14 ; ta có : \(1-\dfrac{1}{n^2}=\dfrac{n^2-1}{n^2}=\dfrac{\left(n-1\right)\left(n+1\right)}{n^2}\)

Ta có : \(A=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{14^2}\right)\)

\(=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}....\dfrac{13.15}{14^2}\)  \(=\dfrac{\left(1.2.3...13\right)}{\left(2.3.4...14\right)}.\dfrac{\left(3.4.5...15\right)}{\left(2.3.4...14\right)}\)

\(=\dfrac{1}{14}.\dfrac{15}{2}=\dfrac{15}{28}\)

\(A=1+3+3^2+...+3^{2020}\)  \(\Rightarrow3A=3+3^2+3^3+...+3^{2021}\)

\(\Rightarrow2A=3^{2021}-1\)  \(\Rightarrow A=\dfrac{3^{2021}-1}{2}\)

B/t = \(sin^225^o+\dfrac{cos^225^o}{sin^225^o}.sin^225^o-tan55^o.cot55^o\)

= \(sin^225^o+cos^225^o-1=1-1=0\)

\(C=\left(\dfrac{3}{2}:\dfrac{3}{4}:\dfrac{4}{5}:...:\dfrac{99}{100}\right).-\dfrac{1}{30}\)

\(=\left(\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{100}{99}\right).-\dfrac{1}{30}\)

\(=\dfrac{100}{2}.-\dfrac{1}{30}=-\dfrac{5}{3}\)

a) P/t \(\Leftrightarrow x-691=-900\Leftrightarrow x=-209\)

Vậy ...

b) P/t \(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3x-6=0\end{matrix}\right.\)  \(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)

Vậy ... 

Gọi số có 2 c/s cần tìm là : \(\overline{ab}\left(a\ne0\right)\)

Nếu viết thêm c/s 0 vào giữa 2 c/s của nó thì số mới là : \(\overline{a0b}\)

Ta có : \(\overline{a0b}=7\overline{ab}\)  \(\Rightarrow100a+b=7\left(10a+b\right)\)

\(\Rightarrow30a=6b\Rightarrow5a=b\)

\(\Rightarrow b=5;a=1\)

Số cần tìm là : 15 

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(2FeCl_2+Cl_2\underrightarrow{t^o}2FeCl_3\)