a) \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}=\frac{100}{101}\)

b) \(\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{97.99}\)

\(=2\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\right)\)

\(=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=2\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(=2.\frac{32}{99}=\frac{64}{99}\)

\(B=\frac{6}{91}+\frac{8}{273}+\frac{4}{525}-\frac{11}{350}\)

\(=\frac{18}{273}+\frac{8}{273}+\frac{8}{1050}-\frac{33}{1050}\)

\(=\left(\frac{18}{273}+\frac{8}{273}\right)+\left(\frac{8}{1050}-\frac{33}{1050}\right)\)

\(=\frac{4}{42}-\frac{1}{42}\)

\(=\frac{1}{14}\)

\(3\left(2x+3\right)\left(3x-5\right)< 0\)

\(\Rightarrow\left(2x+3\right)\left(3x+5\right)< 0\)

Trường hợp 1: \(\Rightarrow\orbr{\hept{\begin{cases}2x+3< 0\\3x-5>0\end{cases}}}\Rightarrow\orbr{\hept{\begin{cases}x< \frac{-3}{2}\\x>\frac{5}{3}\end{cases}}}\)(Loại)

Trường hợp 2: \(\Rightarrow\orbr{\hept{\begin{cases}2x+3>0\\3x-5< 0\end{cases}}}\Rightarrow\orbr{\hept{\begin{cases}x>\frac{-3}{2}\\x< \frac{5}{3}\end{cases}}}\)

Vậy \(\frac{-3}{2}< x< \frac{5}{3}\) thì \(3\left(2x+3\right)\left(3x-5\right)< 0\)

muốn kiếm sp thì phải đc bạn trên 10sp k còn kiếm gp thì phải đc giáo viên k

=> 3/2 : x = 4,5

=> 3/2 : 4,5 = 1/3

hok tốt nhé bạn

là sao vậy ạ

các bạn giúp mik vs

bài nào bạn 

Bài nào ???

\(\frac{x-2}{3}=\frac{2x+1}{4}\)

\(\Rightarrow4\left(x-2\right)=3\left(2x+1\right)\)

\(\Rightarrow4x-4.2=3.2x+3.1\)

\(\Rightarrow4x-8=6x+3\)

\(\Rightarrow4x-6x=3+8\)

\(\Rightarrow-2x=11\)

\(\Rightarrow x=\frac{-11}{2}\)