\(\left(x-2\right)\left(x+\frac{2}{3}\right)>0\)

\(\Rightarrow-\frac{2}{3}< x< 2\)

\(\text{Vậy...}\)

để\(\left(x-2\right).\left(x+\frac{2}{3}\right)>0\)

thì \(x-2;x+\frac{2}{3}\)cùng dấu

\(\Leftrightarrow\hept{\begin{cases}x-2>0\\x+\frac{2}{3}< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>2\\x< -\frac{2}{3}\end{cases}}}\Leftrightarrow2< x< -\frac{2}{3}\left(loại\right)\)

\(\Leftrightarrow\hept{\begin{cases}x-2< 0\\x+\frac{2}{3}>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 2\\x>-\frac{2}{3}\end{cases}}}\Leftrightarrow-\frac{2}{3}< x< 2\left(tm\right)\)

vậy với \(-\frac{2}{3}< x< 2\)thì\(\left(x-2\right).\left(x+\frac{2}{3}\right)>0\)

goi a la so hoc sinh can tim

4000=<a=>4500

theo de bai ta co

(a-4)la boi chung cua 22,24,32

22=2.11;24=2^3.3;32=2^5

=>a=2^5.3.11=1056

tu de bai ta co 4000=<a=>4500 ta co 1056;2112;3168;4224;5280..

=>a=4224 hoc sinh nho la k

ta co n+2=n-1+3 (vi n-1 chia het cho n-1 )=1 nen de A thuoc Z thi 3 thuoc Z

=>3 chia het cho n-1 => n-1 thuoc uoc cua 3 =>n-1=1=>n=2;n-1=-1=> n=0;n-1=3

=>n=4;n-1=-3=>n=-2

kl : n=4;3;0;-2

cam on ban nguyen xuan tro nhung ko phai n+2 ma la   -n+2

74 nha

\(\left|2x-1\right|-3=11\)

=> \(\left|2x-1\right|=11+3\)

=> \(\left|2x-1\right|=14\)

=> \(\orbr{\begin{cases}2x-1=14\\2x-1=-14\end{cases}}\)

=> \(\orbr{\begin{cases}2x=15\\2x=-13\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{15}{2}\\x=-\frac{13}{2}\end{cases}}\)

\(\left|-2x+3\right|-3=x\)

=> \(\left|-2x+3\right|=x+3\) (Đk: \(x+3\ge0\) => \(x\ge-3\))

=> \(\orbr{\begin{cases}-2x+3=x+3\\-2x+3=-x-3\end{cases}}\)

=> \(\orbr{\begin{cases}-3x=0\\-x=-6\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=6\end{cases}}\) (tm)

đề sai rồi bạn ơi

  mk chịu

  banj phải có thêm một số liệu gì nữa chứ

\(B=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)

\(B=\frac{2}{3×5}+\frac{2}{5×7}+\frac{2}{7×9}+...+\frac{2}{19×21}\)

\(B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{19}-\frac{1}{21}\)

\(B=\frac{1}{3}-\frac{1}{21}\)

\(B=\frac{2}{7}\)

A=\(\frac{1}{3}\)+\(\frac{1}{6}\)+\(\frac{1}{10}\)+\(\frac{1}{15}\)+...+\(\frac{1}{66}\) 

A=\(\frac{1}{1\cdot3}\) +\(\frac{1}{2\cdot3}\) +\(\frac{1}{2\cdot5}\)+...+\(\frac{1}{6\cdot11}\)

A=\(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-\frac{1}{5}+...+\frac{1}{6}-\frac{1}{11}\)

A=\(\frac{1}{1}-\frac{1}{11}\)

=>A=\(\frac{10}{11}\)

B=\(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\) 

2B=\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{19\cdot21}\)

2B=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)

2B=\(\frac{1}{3}-\frac{1}{21}\)

2B=\(\frac{2}{7}\)

B=\(\frac{2}{7}:2\)

=>B=\(\frac{1}{7}\)

\(a,4^{2x-6}=1\)\(\Leftrightarrow2x-6=0\Leftrightarrow2x=6\Rightarrow x=3\)

\(b,2^{2x-1}=16\Rightarrow2^{2x-1}=2^4\)

\(\Rightarrow2x-1=4\Rightarrow2x=5\Rightarrow x=\frac{5}{2}\)

\(c,5< 5^x< 125\Rightarrow5^1< 5^x< 5^3\)\(\Rightarrow1< x< 3\)

\(d,5^{x+1}=125\Rightarrow5^{x+1}=5^3\Rightarrow x+1=3\Rightarrow x=2\)

\(a,4^{2x-6}=1\)

\(4^{2x-6}=4^0\)

\(\Rightarrow2x-6=0\)

\(\Rightarrow2x=6\Leftrightarrow x=3\)

\(b,2^{x-1}=16\)

\(2^{x-1}=2^4\)

\(\Rightarrow x-1=4\)

\(\Rightarrow x=5\)