a) Ta có: \(\left(x-\frac{1}{5}\right).\left(x+\frac{4}{7}\right)>0\)

   + \(\hept{\begin{cases}x-\frac{1}{5}>0\\x+\frac{4}{7}>0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x>\frac{1}{5}\\x>-\frac{4}{7}\end{cases}}\)\(\Rightarrow\)\(x>\frac{1}{5}\)

   + \(\hept{\begin{cases}x-\frac{1}{5}< 0\\x+\frac{4}{7}< 0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x< \frac{1}{5}\\x< -\frac{4}{7}\end{cases}}\)\(\Rightarrow\)\(x< -\frac{4}{7}\)

Vậy \(x>\frac{1}{5}\)hoặc \(x< -\frac{4}{7}\)

b) Ta có: \(\left(x+\frac{2}{3}\right).\left(x+2\right)< 0\)

   + \(\hept{\begin{cases}x+\frac{2}{3}>0\\x+2< 0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x>-\frac{2}{3}\\x< -2\end{cases}}\)\(\Rightarrow\)\(-\frac{2}{3}< x< -2\)( vô lí )

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

Vậy \(-2< x< -\frac{2}{3}\)

Ta có: \(\left(2-\frac45+\frac67\right):\left(\frac43+\frac47-\frac{8}{15}\right)\)

\(=\left(\frac{12}{6}-\frac{12}{15}+\frac{12}{14}\right):\left(\frac86+\frac{8}{14}-\frac{8}{15}\right)\)

\(=\frac{12\left(\frac16-\frac{1}{15}+\frac{1}{14}\right)}{8\left(\frac16-\frac{1}{15}+\frac{1}{14}\right)}=\frac{12}{8}=\frac32\)

Ta có: \(\left(\frac{30}{11}-\frac53-\frac{15}{6}\right):\left(\frac{12}{11}-\frac69-1\right)\)

\(=\left(\frac{30}{11}-\frac{30}{18}-\frac{30}{12}\right):\left(\frac{12}{11}-\frac{12}{18}-\frac{12}{12}\right)\)

\(=\frac{30\left(\frac{1}{11}-\frac{1}{18}-\frac{1}{12}\right)}{12\left(\frac{1}{11}-\frac{1}{18}-\frac{1}{12}\right)}=\frac{30}{12}=\frac52\)

Ta có: \(A=\left(2-\frac45+\frac67\right):\left(\frac43+\frac47-\frac{8}{15}\right)-\left(\frac{30}{11}-\frac53-\frac{15}{6}\right):\left(\frac{12}{11}-\frac69-1\right)\)

\(=\frac32-\frac52=-\frac22=-1\)

1) \(\left(\dfrac{-13}{17}-\dfrac{31}{52}\right)-\left(\dfrac{73}{52}-\dfrac{13}{17}+\dfrac{5}{6}\right)-\dfrac{3}{4}\)

\(=\dfrac{-13}{17}-\dfrac{31}{52}-\dfrac{73}{52}+\dfrac{13}{17}-\dfrac{5}{6}-\dfrac{3}{4}\)

\(=\left(\dfrac{-13}{17}+\dfrac{13}{17}\right)-\left(\dfrac{31}{52}+\dfrac{73}{52}\right)-\left(\dfrac{5}{6}+\dfrac{3}{4}\right)\)

\(=0-2-\dfrac{19}{12}\)

\(=-2-\dfrac{19}{12}\)

\(=\dfrac{-43}{12}\)

2) \(\dfrac{1}{7}.\dfrac{1}{3}+\dfrac{1}{7}.\dfrac{1}{2}-\dfrac{1}{7}\)

\(=\dfrac{1}{7}\left(\dfrac{1}{3}+\dfrac{1}{2}-1\right)\)

\(=\dfrac{1}{7}.-\dfrac{1}{6}\)

\(=-\dfrac{1}{42}\)

a: \(=\dfrac{4}{7}+\dfrac{3}{7}\cdot\dfrac{-2}{3}\)

\(=\dfrac{4}{7}-\dfrac{2}{7}=\dfrac{2}{7}\)

x O y t A m C B

a/

\(\widehat{xOt}=\widehat{tOy}=\dfrac{\widehat{xOy}}{2}=\dfrac{60^o}{2}=30^o\)

b/

\(\widehat{xAm}=\widehat{xOy}=60^o\) 

Hai góc trên ở vị trí đồng vị => Am//Oy

c/

Ta có

Am//Oy (cmt) \(\Rightarrow\widehat{ACO}=\widehat{tOy}\) (góc so le trong)

BC//Ox (gt) \(\Rightarrow\widehat{BCO}=\widehat{xOt}\) (góc so le trong)

Mà \(\widehat{xOt}=\widehat{tOy}\left(cmt\right)\)

\(\Rightarrow\widehat{ACO}=\widehat{BCO}\)