2.| 2x - 3 | = \(\frac{1}{2}\)

 | 2x - 3 | = \(\frac{1}{2}:2\)

| 2x - 3 | = \(\frac{1}{4}\)

Th 1 : 2x - 3 = \(\frac{1}{4}\)

        2x = \(\frac{1}{4}+3\)

        2x = \(\frac{13}{4}\)

    x = \(\frac{13}{4}:2\)

x = \(\frac{13}{8}\)

  2 . | 2x - 3 | = 1/2
<=> | 2x - 3 | = 1/4
<=> 2x - 3 = 1/4
hoặc 2x - 3 = -1/4
<=> x = 13/8
hoặc x = 11/8
7,5 - 3 . | 5- xx | = - 4,5
<=> - 3 | 5 - x | = -12
<=> | 5 - x | = 4
<=> 5 - x = 4
hoặc 5 -x = -4
<=> x = 1 hoặc x = 9

Vì AOB là góc bẹt => AOB =180*

Có BOC + COA=AOB

=>BOC=180-50=130*

b,Từ O kẻ 1 tia OM vuông góc vơi AB

Ta có BOD+DOM+MOC+COA=AOB

<=> 40* + DOC +50* = 180* (VÌ DOM+MOC=DOC)

=> DOC =90* hay OC vuông góc với OD

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