a) Ta có: \(\dfrac{2}{3}x-1=\dfrac{3}{2}\)

\(\Leftrightarrow x\cdot\dfrac{2}{3}=\dfrac{5}{2}\)

hay \(x=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{5}{2}\cdot\dfrac{3}{2}=\dfrac{15}{4}\)

b) Ta có: \(\left|5x-\dfrac{1}{2}\right|-\dfrac{2}{7}=25\%\)

\(\Leftrightarrow\left|5x-\dfrac{1}{2}\right|=\dfrac{1}{4}+\dfrac{2}{7}=\dfrac{15}{28}\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{1}{2}=\dfrac{15}{28}\\5x-\dfrac{1}{2}=\dfrac{-15}{28}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{29}{28}\\5x=\dfrac{-1}{28}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{140}\\x=\dfrac{-1}{140}\end{matrix}\right.\)

c) Ta có: \(\dfrac{x-3}{4}=\dfrac{16}{x-3}\)

\(\Leftrightarrow\left(x-3\right)^2=64\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=8\\x-3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\)

d) Ta có: \(\dfrac{-8}{13}+\dfrac{7}{17}+\dfrac{21}{31}\le x\le\dfrac{-9}{14}+4-\dfrac{5}{14}\)

\(\Leftrightarrow\dfrac{3246}{6851}\le x\le3\)

\(\Leftrightarrow x\in\left\{1;2;3\right\}\)

ko biết =))

a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)

b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)

c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)

d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)

e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)

f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)

\(\dfrac{5}{15}+\dfrac{14}{25}-\dfrac{4}{3}+\dfrac{11}{25}\\ =\left(\dfrac{5}{15}-\dfrac{4}{3}\right)+\left(\dfrac{14}{25}+\dfrac{11}{25}\right)\\ =\left(\dfrac{5}{15}-\dfrac{20}{15}\right)+\dfrac{25}{25}\\ =\dfrac{-15}{15}+\dfrac{25}{25}\\ =-1+1\\ =0\)

\(\dfrac{5}{15}\) +  \(\dfrac{14}{25}\) − \(\dfrac{4}{3}\) + \(\dfrac{11}{25}\)\(\dfrac{11}{25}\)

=   [ \(\dfrac{5}{15}\)− \(\dfrac{4}{3}\) ] + [ \(\dfrac{14}{25}\) + 

515+1425−43+1125=(515−43)+(1425+1125)=(515−2015)+2525=−1515+2525=−1+1=0

\(1,\left(dk:x\ne0,-1,4\right)\)

\(\Leftrightarrow\dfrac{9}{x+1}+\dfrac{2}{x-4}-\dfrac{11}{x}=0\)

\(\Leftrightarrow\dfrac{9x\left(x-4\right)+2x\left(x+1\right)-11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=0\)

\(\Leftrightarrow9x^2-36x+2x^2+2x-11x^2+44x-11x+44=0\)

\(\Leftrightarrow-x=-44\)

\(\Leftrightarrow x=44\left(tm\right)\)

\(2,\left(đk:x\ne4\right)\)

\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{2+x}{x-4}-\dfrac{3}{2\left(x-4\right)}+\dfrac{5}{6}=0\)

\(\Leftrightarrow\dfrac{14.2-6\left(2+x\right)-3.3+5\left(x-4\right)}{6\left(x-4\right)}=0\)

\(\Leftrightarrow28-12-6x-9+5x-20=0\)

\(\Leftrightarrow-x=13\)

\(\Leftrightarrow x=-13\left(tm\right)\)

a)= \(\left(\dfrac{4}{9}-\dfrac{17}{18}\right)+\left(\dfrac{17}{14}-\dfrac{5}{7}\right)+\dfrac{11}{125}\)

= \(\dfrac{-1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{11}{125}\)

= 0 + \(\dfrac{11}{125}\)

= \(\dfrac{11}{125}\)

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

\(\left(\dfrac{-2}{3}-\dfrac{1}{3}\right)+\left(3-3\right)+\left(\dfrac{-3}{4}-\dfrac{1}{4}\right)\) + 4

= 0 + (-1) + 0 + (-1) + 0 + (-1) + 4

= -1

c) = \(\dfrac{1}{3}.\dfrac{14}{25}-\dfrac{1}{2}.\dfrac{14}{25}\)

= \(\dfrac{14}{25}.\left(\dfrac{1}{3}-\dfrac{1}{2}\right)\)

= \(\dfrac{14}{25}.\left(\dfrac{-1}{6}\right)\)

= \(\dfrac{-7}{75}\)

d) = \(\left(\dfrac{3}{7}+\dfrac{4}{7}\right)+\left(\dfrac{5}{13}-\dfrac{18}{13}\right)\)

= 1 + (-1)

= 0

a, \(\begin{array}{l}A = \dfrac{{ - 3}}{{14}} + \dfrac{2}{{13}} + \dfrac{{ - 25}}{{14}} + \dfrac{{ - 15}}{{13}}\\A = \left( {\dfrac{{ - 3}}{{14}} + \dfrac{{ - 25}}{{14}}} \right) + \left( {\dfrac{2}{{13}} + \dfrac{{ - 15}}{{13}}} \right)\\A = \dfrac{{ - 3 + \left( { - 25} \right)}}{{14}} + \dfrac{{2 + \left( { - 15} \right)}}{{13}}\\A = \dfrac{{ - 28}}{{14}} + \dfrac{{ - 13}}{{13}}\\A =  - 2 + (-1)\\A =  - 3\end{array}\)

b, 

Cách 1:

\(\begin{array}{l}B = \dfrac{5}{3}.\dfrac{7}{{25}} + \dfrac{5}{3}.\dfrac{{21}}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}\\B = \left( {\dfrac{5}{3}.\dfrac{7}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}} \right) + \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = 0 + \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = \dfrac{{5.21}}{{3.25}}\\B = \dfrac{7}{5}\end{array}\)

Cách 2:

\(B = \dfrac{5}{3}.\dfrac{7}{{25}} + \dfrac{5}{3}.\dfrac{{21}}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}\\B = \dfrac{5}{3}.({\dfrac{7}{{25}} -\dfrac{7}{{25}} + \dfrac{{21}}{{25}}})\\B =  \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = \dfrac{{5.21}}{{3.25}}\\B = \dfrac{7}{5}\)

a) \(\dfrac{5+x}{4-x}=\dfrac{1}{2}\)

\(\Leftrightarrow2\left(5+x\right)=4-x\)

\(\Leftrightarrow2\left(5+x\right)-\left(4-x\right)=0\)

\(\Leftrightarrow10+2x-4+x=0\)

\(\Leftrightarrow6+3x=0\)

\(\Leftrightarrow3x=-6\)

\(\Leftrightarrow x=-2\)

Vậy x=-2

b) \(\dfrac{25}{14}=\dfrac{x+7}{x-4}\)

\(\Leftrightarrow25\left(x-4\right)=14\left(x+7\right)\)

\(\Leftrightarrow25\left(x-4\right)-14\left(x+7\right)=0\)

\(\Leftrightarrow25x-100-14x-98=0\)

\(\Leftrightarrow11x-198=0\)

\(\Leftrightarrow11x=198\)

\(\Leftrightarrow x=18\)

Vậy x=18

c) \(\dfrac{3x-5}{x+4}=\dfrac{5}{2}\)

\(\Leftrightarrow2\left(3x-5\right)=5\left(x+4\right)\)

\(\Leftrightarrow2\left(3x-5\right)-5\left(x+4\right)=0\)

\(\Leftrightarrow6x-10-5x-20=0\)

\(\Leftrightarrow x-30=0\)

\(\Leftrightarrow x=30\)

Vậy x=30

d) \(\dfrac{3x-1}{2x+1}=\dfrac{3}{7}\)

\(\Leftrightarrow7\left(3x-1\right)=3\left(2x+1\right)\)

\(\Leftrightarrow7\left(3x-1\right)-3\left(2x+1\right)=0\)

\(\Leftrightarrow21x-7-6x-3=0\)

\(\Leftrightarrow15x-10=0\)

\(\Leftrightarrow15x=10\)

\(\Leftrightarrow x=\dfrac{10}{15}=\dfrac{2}{3}\)

Vậy \(x=\dfrac{2}{3}\)

5: \(=\dfrac{1}{x-y}\cdot x^3\cdot\left(x-y\right)^2=x^3\left(x-y\right)\)