\(I=\dfrac{5}{4}+\dfrac{-1}{3}-\dfrac{5}{-24}=\dfrac{9}{8}\)

\(J=\dfrac{-19}{-9}+\dfrac{4}{-11}-\dfrac{-2}{3}=\dfrac{239}{99}\)

\(K=\dfrac{-5}{6}-\dfrac{7}{12}+\dfrac{-3}{4}=-\dfrac{13}{6}\)

\(L=\dfrac{-3}{20}+\dfrac{1}{5}-\dfrac{-5}{3}=\dfrac{103}{60}\)

a: =>11(x-3)=6(x-5)

=>11x-33=6x-30

=>5x=3

=>x=3/5

b: =>(4/3-1/4x-5/12)-2x=8/5*5/3=8/3

=>-9/4x+11/12=8/3

=>-9/4x=32/12-11/12=21/12=7/4

=>x=-7/9

c: =>1/2x-1/3-2/3x-1=x

=>-1/6x-4/3=x

=>-7/6x=4/3

=>x=-4/3:7/6=-4/3*6/7=-24/21=-8/7

d: =>1-2x-3x+1=7/2

=>-5x=3/2

=>x=-3/10

a,=\(\dfrac{\left(2-\dfrac{1}{3}+\dfrac{1}{4}\right).12}{\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right).12}\)+\(\dfrac{\left(\dfrac{3}{5}-\dfrac{1}{4}+\dfrac{1}{2}\right).20}{\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{2}{5}\right).20}\)

=\(\dfrac{24-4+3}{24+2-3}\) +\(\dfrac{12-5+10}{10+15-8}\)(nhân từng số hạng với 12;20)

=\(\dfrac{23}{23}\)+\(\dfrac{17}{17}\) =1+1=2

b,=\(\dfrac{5.\left(\dfrac{1}{79}\right)+5.\left(\dfrac{1}{83}\right)+\dfrac{1}{17}}{17.\left(\dfrac{1}{79}\right)+17.\left(\dfrac{1}{83}\right)+\dfrac{1}{5}}\)=\(\dfrac{5.\left(\dfrac{1}{79}+\dfrac{1}{83}\right)+\dfrac{1}{17}}{17.\left(\dfrac{1}{79}+\dfrac{1}{83}\right)+\dfrac{1}{5}}\)

\(A=\dfrac{5}{9}.\dfrac{22}{-3}+\dfrac{5}{9}.\dfrac{5}{-3}=\dfrac{5}{9}\left(\dfrac{22}{-3}+\dfrac{5}{-3}\right)=\dfrac{5}{9}.\dfrac{22+5}{-3}=\dfrac{5}{9}.\dfrac{27}{-3}\)

\(A=\dfrac{5.27}{9.-3}=\dfrac{5.27}{-27}=-5\)

\(B=\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)

\(B=\left(6-5-3\right)+\left(\dfrac{-2}{3}-\dfrac{5}{3}+\dfrac{7}{3}\right)+\left(\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\right)\)

\(B=-2+\left(\dfrac{-2-5+7}{2}\right)+\left(\dfrac{1+3-5}{2}\right)\)

\(B=-2+\dfrac{0}{2}+\left(\dfrac{-1}{2}\right)=-2+0-\dfrac{1}{2}=\dfrac{-5}{2}\)

a, \(\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{2}{5}\right)>0\)

\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{3}>0\\x+\dfrac{2}{5}>0\end{matrix}\right.\) hay \(\left\{{}\begin{matrix}x-\dfrac{1}{3}< 0\\x+\dfrac{2}{5}< 0\end{matrix}\right.\)

+,Xét \(\left\{{}\begin{matrix}x-\dfrac{1}{3}>0\\x+\dfrac{2}{5}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>\dfrac{1}{3}\\x>-\dfrac{2}{5}\end{matrix}\right.\)

\(\Rightarrow x>\dfrac{1}{3}\)

+, Xét \(\left\{{}\begin{matrix}x-\dfrac{1}{3}< 0\\x+\dfrac{2}{5}< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< \dfrac{1}{3}\\x< -\dfrac{2}{5}\end{matrix}\right.\)

\(\Rightarrow x< -\dfrac{2}{5}\)

Vậy...........

b, \(\left(x+\dfrac{3}{5}\right)\left(x+1\right)< 0\)

Vì \(x+\dfrac{3}{5}< x+1\) với mọi \(x\in R\)

\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{5}< 0\\x+1>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< -\dfrac{3}{5}\\x>-1\end{matrix}\right.\)

Vậy...........

c, \(\dfrac{3}{7}x-\dfrac{2}{5}x=\dfrac{-17}{35}\)

\(\Rightarrow\dfrac{1}{35}x=\dfrac{-17}{35}\)

\(\Rightarrow x=-17\)

d, \(\left(\dfrac{3}{4}x-\dfrac{9}{10}\right)\left(\dfrac{1}{3}+\dfrac{-3}{5}x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{9}{10}=0\\\dfrac{1}{3}+\dfrac{-3}{5}x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{9}{10}\\-\dfrac{3}{5}x=-\dfrac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=\dfrac{5}{9}\end{matrix}\right.\)

Vậy.........

Chúc bạn học tốt!!!

a/ \(\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{2}{5}\right)>0\)

TH1:\(\left\{{}\begin{matrix}x-\dfrac{1}{3}>0\\x+\dfrac{2}{5}>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>\dfrac{1}{3}\\x>-\dfrac{2}{5}\end{matrix}\right.\)\(\Rightarrow x>\dfrac{1}{3}\)

TH2:\(\left\{{}\begin{matrix}x-\dfrac{1}{3}< 0\\x+\dfrac{2}{5}< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< \dfrac{1}{3}\\x< -\dfrac{2}{5}\end{matrix}\right.\)\(\Rightarrow x< -\dfrac{2}{5}\)

Vậy \(x>\dfrac{1}{3}\) hoặc \(x< -\dfrac{2}{5}\) thì tm

b/ \(\left(x+\dfrac{3}{5}\right)\left(x+1\right)< 0\)

TH1:\(\left\{{}\begin{matrix}x+\dfrac{3}{5}< 0\\x+1>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< -\dfrac{3}{5}\\x>-1\end{matrix}\right.\) \(\Rightarrow-1< x< -\dfrac{3}{5}\)

TH2:\(\left\{{}\begin{matrix}x+\dfrac{3}{5}>0\\x+1< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>-\dfrac{3}{5}\\x< -1\end{matrix}\right.\)(vô lý)

Vậy....................

c/ \(\dfrac{3}{7}x-\dfrac{2}{5}x=-\dfrac{17}{35}\)

\(\Rightarrow\left(\dfrac{3}{7}-\dfrac{2}{5}\right)x=-\dfrac{17}{35}\)

\(\Rightarrow\dfrac{1}{35}x=-\dfrac{17}{35}\)

\(\Rightarrow x=-\dfrac{17}{35}:\dfrac{1}{35}=-17\)

Vậy.............

d/ \(\left(\dfrac{3}{4}x-\dfrac{9}{10}\right)\left(\dfrac{1}{3}+\dfrac{-3}{5}x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{9}{10}=0\\\dfrac{1}{3}-\dfrac{3}{5}x=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{9}{10}\\\dfrac{3}{5}x=\dfrac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=\dfrac{5}{9}\end{matrix}\right.\)

Vậy.....................

A= 3/5.6/7+3/5.3/7-3/5.2/7

A= 3/5. (6/7+3/7-2/7) = 3/5

B= (-13.2/5+-2/9.2/5+2/5.11/9).5/2

B= (-13-2/9+11/9).2/5.5/2

B= -13+(11/9-2/9)= -12

C= (-4/5+5/7).3/2+(-1/5+2/7).3/2

C= (-4/5+5/7+-1/5+2/7).3/2

C= ((-4/5+-1/5)+(5/7+2/7)).3/2

C= 0

D= 4/9.-5/3+4/9.-22/3

D= 4/9.110/9

D= 440/81

.

a, \(\dfrac{3}{5}-4.\left|\dfrac{1}{5}-\dfrac{3}{4}x\right|=\dfrac{1}{3}\)

\(\Rightarrow4\left|\dfrac{1}{5}-\dfrac{3}{4}x\right|=\dfrac{4}{15}\)

\(\Rightarrow\left|\dfrac{1}{5}-\dfrac{3}{4}x\right|=\dfrac{1}{15}\)

\(\Rightarrow\dfrac{1}{5}-\dfrac{3}{4}x\in\left\{-\dfrac{1}{15};\dfrac{1}{15}\right\}\)

\(\Rightarrow\dfrac{3}{4}x\in\left\{\dfrac{4}{15};\dfrac{2}{15}\right\}\Rightarrow x\in\left\{\dfrac{16}{45};\dfrac{8}{45}\right\}\)

b, \(\left|2\dfrac{2}{9}-x\right|=\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\)

\(\Rightarrow\left|2\dfrac{2}{9}-x\right|=\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}\)

\(\Rightarrow\left|2\dfrac{2}{9}-x\right|=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{8}-\dfrac{1}{9}\)

(do \(\dfrac{1}{a.\left(a+1\right)}=\dfrac{1}{a}-\dfrac{1}{a+1}\) với mọi \(a\in N\)*)

\(\Rightarrow\left|2\dfrac{2}{9}-x\right|=\dfrac{1}{3}-\dfrac{1}{9}\)

\(\Rightarrow\left|2\dfrac{2}{9}-x\right|=\dfrac{2}{9}\Rightarrow2\dfrac{2}{9}-x\in\left\{-\dfrac{2}{9};\dfrac{2}{9}\right\}\)

\(\Rightarrow x\in\left\{\dfrac{22}{9};2\right\}\)

c,\(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\)

\(\Rightarrow\dfrac{1}{3}x+\dfrac{2}{5}x-\dfrac{2}{5}=0\)

\(\Rightarrow\dfrac{11}{15}x=\dfrac{2}{5}\Rightarrow x=\dfrac{6}{11}\)

d, \(60\%x+\dfrac{2}{3}x=\dfrac{1}{3}.6\dfrac{1}{3}\)

\(\Rightarrow\dfrac{3}{5}x+\dfrac{2}{3}x=\dfrac{1}{3}.\dfrac{19}{3}\)

\(\Rightarrow\dfrac{19}{15}x=\dfrac{19}{9}\Rightarrow x=\dfrac{5}{3}\)

Chúc bạn học tốt!!!

\(\dfrac{3}{8}\)+2²-\(\dfrac{3}{8}\)

=(\(\dfrac{3}{8}\)-\(\dfrac{3}{8}\))+2² (2² =4)

=4

b, \(\dfrac{2}{5}\).33\(\dfrac{1}{3}\)-\(\dfrac{2}{5}\).\(8\dfrac{1}{3}\)

=\(\dfrac{2}{5}\).\(\dfrac{100}{3}\)-\(\dfrac{2}{5}\).\(\dfrac{25}{3}\)

=(\(\dfrac{2}{5}\)-\(\dfrac{2}{5}\)).\(\dfrac{100}{3}\).\(\dfrac{25}{3}\)

=0

\(\dfrac{1}{2-\dfrac{3}{4+\dfrac{5}{6-\dfrac{7}{8+\dfrac{9}{10}}}}}=\dfrac{1}{x+\dfrac{1}{3+\dfrac{1}{5}}}+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{2}}}\\ \Leftrightarrow\dfrac{767}{1070}=\dfrac{1}{x+\dfrac{5}{16}}+\dfrac{3}{5}\\ \Leftrightarrow\dfrac{25}{214}=\dfrac{1}{x+\dfrac{5}{16}}\\ \Rightarrow x+\dfrac{5}{16}=\dfrac{214}{25}\Rightarrow x=\dfrac{3299}{400}\)