tìm số đối của các tổng sau :
a) \(\dfrac{1}{-2}\) + \(\dfrac{2}{3}\)
b) \(\dfrac{-3}{4}\) + \(\dfrac{-4}{3}\)
c) \(\dfrac{7}{-2}\) + \(\dfrac{-3}{4}\)
d)-2 + \(\dfrac{3}{-4}\)
Bài 1: Quy đồng mẫu số các phân số sau
a) \(\dfrac{1}{2}\);\(\dfrac{2}{3}\) và \(\dfrac{3}{4}\)
b) \(\dfrac{1}{3}\);\(\dfrac{2}{15}\) và \(\dfrac{4}{45}\)
c) \(\dfrac{1}{8}\);\(\dfrac{2}{3}\) và \(\dfrac{5}{2}\)
d) \(\dfrac{2}{7}\);\(\dfrac{9}{4}\) và \(\dfrac{5}{28}\)
(Các bạn ko cần viết kết luận đâu ah!)
Bài 2:
a) Hãy viết 4 và \(\dfrac{9}{4}\) thành hai phân số có mẫu số chung là 12.
b) Hãy viết \(\dfrac{5}{8}\);\(\dfrac{25}{30}\) và 2 thành các phân số có mẫu số chung là 240.
Bài 1:
a)
\(\dfrac{1}{2}=\dfrac{1\times6}{2\times6}=\dfrac{6}{12}\)
\(\dfrac{2}{3}=\dfrac{2\times4}{3\times4}=\dfrac{8}{12}\)
\(\dfrac{3}{4}=\dfrac{3\times3}{4\times3}=\dfrac{9}{12}\)
b)
\(\dfrac{1}{3}=\dfrac{1\times15}{3\times15}=\dfrac{15}{45}\)
\(\dfrac{2}{15}=\dfrac{2\times3}{15\times3}=\dfrac{6}{45}\)
\(\dfrac{4}{45}\) (giữ nguyên)
c)
\(\dfrac{1}{8}=\dfrac{1\times3}{8\times3}=\dfrac{3}{24}\)
\(\dfrac{2}{3}=\dfrac{2\times8}{3\times8}=\dfrac{16}{24}\)
\(\dfrac{5}{2}=\dfrac{5\times12}{2\times12}=\dfrac{60}{24}\)
d)
\(\dfrac{2}{7}=\dfrac{2\times4}{7\times4}=\dfrac{8}{28}\)
\(\dfrac{9}{4}=\dfrac{9\times7}{4\times7}=\dfrac{63}{28}\)
\(\dfrac{5}{28}\) (giữ nguyên)
Bài 2:
a)
\(4=\dfrac{4}{1}=\dfrac{4\times12}{1\times12}=\dfrac{48}{12}\)
\(\dfrac{9}{4}=\dfrac{9\times3}{4\times3}=\dfrac{27}{12}\)
b)
\(\dfrac{5}{8}=\dfrac{5\times30}{8\times30}=\dfrac{150}{240}\)
\(\dfrac{25}{30}=\dfrac{5}{6}=\dfrac{5\times40}{6\times40}=\dfrac{200}{240}\)
\(2=\dfrac{2}{1}=\dfrac{2\times240}{1\times240}=\dfrac{480}{240}\).
Tìm x, biết:
a) \(\dfrac{-2}{5}\) + \(\dfrac{4}{5}\) . x = \(\dfrac{3}{5}\)
b) \(\dfrac{-3}{7}\) - \(\dfrac{4}{7}\) : x = \(\dfrac{2}{5}\)
c) \(\dfrac{4}{7}\) . x + \(\dfrac{2}{3}\) = \(\dfrac{-1}{5}\)
d) \(\dfrac{5}{7}\) : x -1 = \(\dfrac{2}{3}\)
a, - \(\dfrac{2}{5}\) + \(\dfrac{4}{5}\).\(x\) = \(\dfrac{3}{5}\)
\(\dfrac{4}{5}\).\(x\) = \(\dfrac{3}{5}\)+ \(\dfrac{2}{5}\)
\(\dfrac{4}{5}\).\(x\) = 1
\(x\) = \(\dfrac{5}{4}\)
b, - \(\dfrac{3}{7}\) - \(\dfrac{4}{7}\): \(x\) = \(\dfrac{2}{5}\)
\(\dfrac{4}{7}\): \(x\) = - \(\dfrac{3}{7}\) - \(\dfrac{2}{5}\)
\(\dfrac{4}{7}\): \(x\) = - \(\dfrac{29}{35}\)
\(x\) = \(\dfrac{4}{7}\): (- \(\dfrac{29}{35}\) )
\(x\) = - \(\dfrac{20}{29}\)
c, \(\dfrac{4}{7}\).\(x\) + \(\dfrac{2}{3}\) = - \(\dfrac{1}{5}\)
\(\dfrac{4}{7}\).\(x\) = -\(\dfrac{1}{5}\) - \(\dfrac{2}{3}\)
\(\dfrac{4}{7}\).\(x\) = - \(\dfrac{13}{15}\)
\(x\) = - \(\dfrac{13}{15}\): \(\dfrac{4}{7}\)
\(x\) = - \(\dfrac{91}{60}\)
d, \(\dfrac{5}{7}\): \(x\) - 1 = \(\dfrac{2}{3}\)
\(\dfrac{5}{7}\): \(x\) = \(\dfrac{2}{3}\)+ 1
\(\dfrac{5}{7}\): \(x\) = \(\dfrac{5}{3}\)
\(x\) = \(\dfrac{5}{7}\): \(\dfrac{5}{3}\)
\(x\) = \(\dfrac{3}{7}\)
Tìm x, biết:
a) \(\dfrac{2}{5}\) + \(\dfrac{3}{4}\): x = \(\dfrac{-1}{2}\)
b) \(\dfrac{5}{7}\) - \(\dfrac{2}{3}\) . x = \(\dfrac{4}{5}\)
c) \(\dfrac{1}{2}\) x + \(\dfrac{2}{3}\) x = \(\dfrac{-2}{3}\)
d) \(\dfrac{4}{7}\)x - x= \(\dfrac{-9}{14}\)
a, 2/5 + 3/4 : x = -1/2
3/4 : x = -1/2 - 2/5
3/4 : x = -9/10
x = 3/4 : -9/10
x = -5/6
b, 5/7 - 2/3 . x = 4/5
2/3 . x = 4/5 + 5/7
2/3 . x = 53/35
x = 53/35 : 2/3
x = 159/70
c và d mình làm dược nhưng ko ghi được cái suy ra
Tìm x biết: a) x + \(\dfrac{2}{3}=\dfrac{4}{27}\) b) \(\dfrac{3}{4}x-\dfrac{7}{3}=\dfrac{1}{4}x+\dfrac{1}{6}\)
c) \(\dfrac{13}{10}x-\dfrac{5}{2}=\dfrac{7}{2}\) d) (3\(x\) + 2) \(\left(\dfrac{-2}{5}x-7\right)=0\)
a: x=4/27-2/3=4/27-18/27=-14/27
b: =>3/4x-1/4x=1/6+7/3
=>1/2x=1/6+14/6=5/2
hay x=5
c: =>13/10x=7/2+5/2=6
=>x=13/10:6=13/60
d: (3x+2)(-2/5x-7)=0
=>3x+2=0 hoặc 2/5x+7=0
=>x=-2/3 hoặc x=-35/2
a: x=4/27-2/3=4/27-18/27=-14/27
b: =>3/4x-1/4x=1/6+7/3
=>1/2x=1/6+14/6=5/2
hay x=5
c: =>13/10x=7/2+5/2=6
=>x=13/10:6=13/60
d: (3x+2)(-2/5x-7)=0
=>3x+2=0 hoặc 2/5x+7=0
=>x=-2/3 hoặc x=-35/2
a) \(\dfrac{2}{5}+\dfrac{4}{3}:\dfrac{-2}{3}\)
b) \(3\dfrac{4}{5}-\left(2\dfrac{1}{4}+1\dfrac{4}{5}\right)\)
c) \(\dfrac{-3}{5}.\dfrac{4}{7}+\dfrac{3}{7}.\dfrac{-3}{5}+\dfrac{3}{5}\)
d) 40% \(-1\dfrac{5}{7}:\dfrac{3}{7}+\left|\dfrac{-9}{5}\right|\)
\(a.\dfrac{2}{3}+\dfrac{4}{3}:\dfrac{-2}{3}=\dfrac{2}{3}+\left(-2\right)=\dfrac{-4}{3}\)
\(b.3\dfrac{4}{5}-\left(2\dfrac{1}{4}+1\dfrac{4}{5}\right)\\ =3\dfrac{4}{5}-2\dfrac{1}{4}-1\dfrac{4}{5}\\ =\left(3\dfrac{4}{5}-1\dfrac{4}{5}\right)-2\dfrac{1}{4}\\ =2-2\dfrac{1}{4}=\dfrac{1}{4}\)
\(c.\dfrac{-3}{5}.\dfrac{4}{7}+\dfrac{3}{7}.\dfrac{-3}{5}+\dfrac{3}{5}\\ =\dfrac{-3}{5}\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\dfrac{3}{5}\\ =\dfrac{-3}{5}+\dfrac{3}{5}=0\)
a) \(\dfrac{2}{5}+\dfrac{4}{3}:\dfrac{-2}{3}\)
\(=\dfrac{2}{5}+\dfrac{4}{3}.\dfrac{-3}{2}\)
\(=\dfrac{2}{5}+-2\)
\(=\dfrac{2}{5}+\dfrac{-10}{5}\)
\(=\dfrac{-8}{5}\)
a) \(\dfrac{2}{5}+\dfrac{4}{3}:\dfrac{-2}{3}=\dfrac{2}{5}+-2=\dfrac{-8}{5}\)
b) \(3\dfrac{4}{5}-\left(2\dfrac{1}{4}+1\dfrac{4}{5}\right)=\dfrac{19}{5}-\dfrac{9}{4}-\dfrac{9}{5}=\left(\dfrac{19}{5}-\dfrac{9}{5}\right)-\dfrac{9}{4}=2-\dfrac{9}{4}=\dfrac{-1}{4}\)
c) \(\dfrac{-3}{5}.\dfrac{4}{7}+\dfrac{3}{7}.\dfrac{-3}{5}+\dfrac{3}{5}\)
\(=\dfrac{-3}{5}.\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\dfrac{3}{5}\)
\(=\dfrac{-3}{5}.1+\dfrac{3}{5}\)
\(=\dfrac{-3}{5}+\dfrac{3}{5}\)
\(=0\)
d) \(40\%-1\dfrac{5}{7}:\dfrac{3}{7}+\left|\dfrac{-9}{5}\right|\)
\(=\dfrac{2}{5}-\dfrac{12}{7}:\dfrac{3}{7}+\dfrac{9}{5}\)
\(=\dfrac{2}{5}-4+\dfrac{9}{5}\)
\(=\dfrac{-9}{5}\)
làm đầy đủ theo các bước nhé
Tìm x biết :
a) \(^{\dfrac{4}{9}+x=\dfrac{5}{3}}\)
b)\(\dfrac{3}{4}.x=\dfrac{-1}{2}\)
c) \(\dfrac{3}{7}+\dfrac{5}{7}:x=\dfrac{1}{3}\)
d)\(3\dfrac{1}{4}:\left|2x-\dfrac{5}{12}\right|=\dfrac{39}{16}\)
c.\(\dfrac{3}{7}+\dfrac{5}{7}:x=\dfrac{1}{3}\)
\(\dfrac{5}{7}:x=\dfrac{1}{3}-\dfrac{3}{7}\)
\(\dfrac{5}{7}:x=-\dfrac{2}{21}\)
\(x=\dfrac{5}{7}:-\dfrac{2}{21}\)
\(x=-\dfrac{15}{2}\)
d.\(3\dfrac{1}{4}:\left|2x-\dfrac{5}{12}\right|=\dfrac{39}{16}\)
\(\left|2x-\dfrac{5}{12}\right|=3\dfrac{1}{4}:\dfrac{39}{16}\)
\(\left|2x-\dfrac{5}{12}\right|=\dfrac{4}{3}\)
\(\rightarrow\left[{}\begin{matrix}2x-\dfrac{5}{12}=\dfrac{4}{3}\\2x-\dfrac{4}{12}=-\dfrac{4}{3}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}2x=\dfrac{7}{4}\\2x=-\dfrac{11}{12}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}x=\dfrac{7}{8}\\x=-\dfrac{11}{24}\end{matrix}\right.\)
A, \(\dfrac{4}{9}+x=\dfrac{5}{3}\)
\(x\)\(=\dfrac{5}{3}-\dfrac{4}{9}\)
\(x\)\(=\dfrac{11}{9}\)
B,\(\dfrac{3}{4}.x=\dfrac{-1}{2}\)
\(x=\dfrac{-1}{2}:\dfrac{3}{4}\)
\(x=\)\(\dfrac{-2}{3}\)
a)
\(\frac{4}{9} + x = \frac{5}{3}\)
=> \(x = \frac{5}{3}-\frac{4}{9}\)
=> \(x = \) \(\frac{11}{9}\)
Vậy \(x = \dfrac{11}{9}\)
b)
\(\dfrac{3}{4} .x = \dfrac{-1}{2}\)
=> \(x = \dfrac{-1}{2} : \dfrac{3}{4}\)
=> \(x = \dfrac{-2}{3}\)
Vậy \(x = \dfrac{-2}{3}\)
c)
\( \dfrac{3}{7}+ \dfrac{5}{7}:x = \dfrac{1}{3}\)
=> \(\dfrac{5}{7}:x = \dfrac{1}{3}-\) \( \dfrac{3}{7}\)
=> \(\dfrac{5}{7}:x = \dfrac{-2}{21}\)
=> \(x = \dfrac{5}{7}:\dfrac{-2}{21}\)
=> \(x = \dfrac{-15}{2}\)
Vậy \(x = \dfrac{-15}{2}\)
d)
\(3\dfrac{1}{4} : |2x - \dfrac{5}{12} | = \dfrac{39}{16}\)
=> \(\dfrac{13}{4} : |2x - \dfrac{5}{12} | = \dfrac{39}{16}\)
=> \( |2x - \dfrac{5}{12} | =\dfrac{13}{4} : \dfrac{39}{16}\)
=> \(|2x-\dfrac{5}{12} |= \dfrac{4}{3}\)
=> \(\left[\begin{matrix} 2x - \dfrac{5}{12} = \dfrac{4}{3}\\ 2x - \dfrac{5}{12} = \dfrac{4}{3}\end{matrix}\right.\)
=> \(\left[\begin{matrix} 2x = \dfrac{-4}{3}+\dfrac{5}{12}\\ 2x = \dfrac{-4}{3}+\dfrac{5}{12} \end{matrix}\right.\)
=> \(\left[\begin{matrix} 2x = \dfrac{7}{4}\\ 2x = \dfrac{-11}{12} \end{matrix}\right.\)
=> \(\left[\begin{matrix} x = \dfrac{7}{8}\\ x = \dfrac{-11}{24} \end{matrix}\right.\)
Vậy \(x \in \) { \(\dfrac{7}{8} ; \dfrac{-11}{24}\) }
Tìm x:
a) \(\dfrac{11}{12}\) - (\(\dfrac{2}{5}\) + \(\dfrac{3}{4}\)x) \(\dfrac{2}{3}\)
b) \(\dfrac{-2}{5}\) + \(\dfrac{5}{3}\) . (\(\dfrac{3}{2}\) - \(\dfrac{4}{15}\)x) = \(\dfrac{-7}{6}\)
c) \(\dfrac{1}{2}\) + \(\dfrac{3}{4}\)x = \(\dfrac{1}{4}\)
`#3107`
a)
\(\dfrac{11}{12}-\left(\dfrac{2}{5}+\dfrac{3}{4}x\right)=\dfrac{2}{3}?\\ \Rightarrow\dfrac{2}{5}+\dfrac{3}{4}x=\dfrac{11}{12}-\dfrac{2}{3}\\ \Rightarrow\dfrac{2}{5}+\dfrac{3}{4}x=\dfrac{1}{4}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{1}{4}-\dfrac{2}{5}\\ \Rightarrow\dfrac{3}{4}x=-\dfrac{3}{20}\\ \Rightarrow x=-\dfrac{3}{20}\div\dfrac{3}{4}\\ \Rightarrow x=-\dfrac{1}{5}\)
Vậy, \(x=-\dfrac{1}{5}\)
b)
\(\dfrac{-2}{5}+\dfrac{5}{3}\cdot\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=\dfrac{-7}{6}\\ \Rightarrow\dfrac{5}{3}\cdot\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=\dfrac{-7}{6}-\dfrac{-2}{5}\\ \Rightarrow\dfrac{5}{3}\cdot\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=-\dfrac{23}{30}\\ \Rightarrow\dfrac{3}{2}-\dfrac{4}{15}x=-\dfrac{23}{30}\div\dfrac{5}{3}\\ \Rightarrow\dfrac{3}{2}-\dfrac{4}{15}x=-\dfrac{23}{50}\\ \Rightarrow\dfrac{4}{15}x=\dfrac{3}{2}-\left(-\dfrac{23}{50}\right)\\ \Rightarrow\dfrac{4}{15}x=\dfrac{49}{25}\\ \Rightarrow x=\dfrac{147}{20}\)
Vậy, \(x=\dfrac{147}{20}\)
c)
\(\dfrac{1}{2}+\dfrac{3}{4}x=\dfrac{1}{4}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{1}{4}-\dfrac{1}{2}\\ \Rightarrow\dfrac{3}{4}x=-\dfrac{1}{4}\\ \Rightarrow x=-\dfrac{1}{4}\div\dfrac{3}{4}\\ \Rightarrow x=-\dfrac{1}{3}\)
Vậy, \(x=-\dfrac{1}{3}.\)
\(#Emyeu1aithatroi...\)
(2/5 + 3/4 . x)= 11/12 -2/3
(2/5 +3/4 . x)= 1/4
3/4 . x = 1/4 - 2/5
3/4 . x = -3/20
x = -3/20 : 3/4
x = -1/5
Vậy .....
Tính giá trí của biểu thức sau theo cách hợp lí nhất.
a) $\mathrm{A}=\dfrac{3}{5} \cdot \dfrac{6}{7}+\dfrac{3}{7}: \dfrac{5}{3}-\dfrac{2}{7}: 1 \dfrac{2}{3}$;
b) $\mathrm{B}=\left(-13 \cdot \dfrac{2}{5}+\dfrac{-2}{9}: 2 \dfrac{1}{2}+\dfrac{2}{5}. \dfrac{11}{9}\right) \cdot 2 \dfrac{1}{2}$;
c) $\mathrm{C}=\left(\dfrac{-4}{5}+\dfrac{5}{7}\right): \dfrac{2}{3}+\left(\dfrac{-1}{5}+\dfrac{2}{7}\right): \dfrac{2}{3}$;
d) $\mathrm{D}=\dfrac{4}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)+\dfrac{4}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)$.
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
Bài 2:Tìm x biết:
a)\(\dfrac{1}{7}+x=\dfrac{-2}{3}\)
b)\(\dfrac{-2}{3}:x=\dfrac{-5}{6}\)
c)\(\left\{\dfrac{3}{5}-2x\right\}.\dfrac{5}{8}=1\)
d)\(\dfrac{3}{4}+\dfrac{2}{5}x=\dfrac{29}{60}\)
e)\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
f)\(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
g)\(\left|X+\dfrac{1}{3}\right|-4=\dfrac{-1}{2}\)
h)\(\left(\dfrac{1}{32}\right)^x.8^{2x}=512\)
i)\(5,3x+\left(-3,3\right)x+1,7=-4,9\)
a) Ta có: \(\dfrac{1}{7}+x=-\dfrac{2}{3}\)
\(\Leftrightarrow x=-\dfrac{2}{3}-\dfrac{1}{7}=\dfrac{-14}{21}-\dfrac{3}{21}\)
hay \(x=-\dfrac{17}{21}\)
Vậy: \(x=-\dfrac{17}{21}\)
b) Ta có: \(\dfrac{-2}{3}:x=\dfrac{-5}{6}\)
\(\Leftrightarrow x=\dfrac{-2}{3}:\dfrac{-5}{6}=\dfrac{-2}{3}\cdot\dfrac{6}{-5}=\dfrac{-12}{-15}=\dfrac{4}{5}\)
Vậy: \(x=\dfrac{4}{5}\)
c) Ta có: \(\left(\dfrac{3}{5}-2x\right)\cdot\dfrac{5}{8}=1\)
\(\Leftrightarrow\left(\dfrac{3}{5}-2x\right)=1:\dfrac{5}{8}=\dfrac{8}{5}\)
\(\Leftrightarrow-2x=\dfrac{8}{5}-\dfrac{3}{5}=1\)
hay \(x=-\dfrac{1}{2}\)
Vậy: \(x=-\dfrac{1}{2}\)
d) Ta có: \(\dfrac{3}{4}+\dfrac{2}{5}x=\dfrac{29}{60}\)
\(\Leftrightarrow x\cdot\dfrac{2}{5}=\dfrac{29}{60}-\dfrac{3}{4}=\dfrac{29}{60}-\dfrac{45}{60}=\dfrac{-16}{60}=\dfrac{-4}{15}\)
hay \(x=\dfrac{-4}{15}:\dfrac{2}{5}=\dfrac{-4}{15}\cdot\dfrac{5}{2}=\dfrac{-20}{30}=-\dfrac{2}{3}\)
Vậy: \(x=-\dfrac{2}{3}\)
e) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)
hay \(x=-\dfrac{1}{4}:\dfrac{7}{20}=\dfrac{-1}{4}\cdot\dfrac{20}{7}=\dfrac{-20}{28}=\dfrac{-5}{7}\)
Vậy: \(x=-\dfrac{5}{7}\)
f) Ta có: \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\Leftrightarrow-x+\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}=0\)
\(\Leftrightarrow-x+\dfrac{55}{60}-\dfrac{24}{60}-\dfrac{40}{60}=0\)
\(\Leftrightarrow-x-\dfrac{9}{60}=0\)
\(\Leftrightarrow-x=\dfrac{9}{60}=\dfrac{3}{20}\)
hay \(x=-\dfrac{3}{20}\)
Vậy: \(x=-\dfrac{3}{20}\)
g) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=\dfrac{-1}{2}\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{-1}{2}+4=\dfrac{-1}{2}+\dfrac{8}{2}=\dfrac{7}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{7}{2}\\x+\dfrac{1}{3}=-\dfrac{7}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{21}{6}-\dfrac{2}{6}=\dfrac{19}{6}\\x=-\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{-21}{6}-\dfrac{2}{6}=\dfrac{-23}{6}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{19}{6};-\dfrac{23}{6}\right\}\)