a) \(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\)

\(-\dfrac{5}{6}x=\dfrac{5}{12}\)

\(x=-\dfrac{1}{2}\)

b) \(\dfrac{2}{5}+\dfrac{3}{5}\cdot\left(3x-3.7\right)=-\dfrac{53}{10}\)

\(\dfrac{3}{5}\left(3x-3.7\right)=-\dfrac{57}{10}\)

\(3x-3.7=-\dfrac{19}{2}\)

\(3x=-5.8\)

\(x=-\dfrac{29}{15}\)

c) \(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)+\dfrac{5}{9}=\dfrac{23}{27}\)

\(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)=\dfrac{8}{27}\)

\(2+\dfrac{3}{4}x=\dfrac{21}{8}\)

\(\dfrac{3}{4}x=\dfrac{5}{8}\)

\(x=\dfrac{5}{6}\)

d) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)

\(-\dfrac{2}{3}x=\dfrac{1}{10}\)

\(x=-\dfrac{3}{20}\)

\(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\)

\(\left(\dfrac{2}{3}-\dfrac{3}{2}\right)x=\dfrac{5}{12}\)

\(\dfrac{-5}{6}.x=\dfrac{5}{12}\)

-> x = \(\dfrac{-1}{2}\)

Bài 1:

a)

\(\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{x-1}{9}=\dfrac{24}{9}\\ \Leftrightarrow x-1=24\\ x=24+1\\ x=25\)

b)

\(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{8}\\ \dfrac{3x}{7}+1=\dfrac{-1}{8}\cdot\left(-4\right)\\ \dfrac{3x}{7}+1=\dfrac{1}{2}\\ \dfrac{3x}{7}=\dfrac{1}{2}-1\\ \dfrac{3x}{7}=\dfrac{-1}{2}\\ 3x=\dfrac{-1}{2}\cdot7\\ 3x=\dfrac{-7}{2}\\ x=\dfrac{-7}{2}:3\\ x=\dfrac{-7}{6}\)

c)

\(x+\dfrac{7}{12}=\dfrac{17}{18}-\dfrac{1}{9}\\ x+\dfrac{7}{12}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{12}\\ x=\dfrac{1}{4}\)

d)

\(0,5x-\dfrac{2}{3}x=\dfrac{7}{12}\\ \dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{2}{3}\right)=\dfrac{7}{12}\\ \dfrac{-1}{6}x=\dfrac{7}{12}\\ x=\dfrac{7}{12}:\dfrac{-1}{6}\\ x=\dfrac{-7}{2}\)

e)

\(\dfrac{29}{30}-\left(\dfrac{13}{23}+x\right)=\dfrac{7}{46}\\ \dfrac{29}{30}-\dfrac{13}{23}-x=\dfrac{7}{46}\\ \dfrac{277}{690}-x=\dfrac{7}{46}\\ x=\dfrac{277}{690}-\dfrac{7}{46}\\ x=\dfrac{86}{345}\)

f)

\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\\ \left(x-\dfrac{1}{12}\right):\dfrac{23}{12}=\dfrac{7}{46}\\ x-\dfrac{1}{12}=\dfrac{7}{46}\cdot\dfrac{23}{12}\\ x-\dfrac{1}{12}=\dfrac{7}{24}\\ x=\dfrac{7}{24}+\dfrac{1}{12}\\ x=\dfrac{3}{8}\)

g)

\(\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{1}{6}\\ \dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\\ \dfrac{13}{21}+x=\dfrac{2}{7}\\ x=\dfrac{2}{7}-\dfrac{13}{21}\\ x=\dfrac{-1}{3}\)

h)

\(2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\end{matrix}\right.\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\ \dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{29}{24}\\ x=\dfrac{29}{24}:\dfrac{1}{2}\\ x=\dfrac{29}{12}\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\\ \dfrac{1}{2}x=\dfrac{-7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{-13}{24}\\ x=\dfrac{-13}{24}:\dfrac{1}{2}\\ x=\dfrac{-13}{12}\)

i)

\(3\cdot\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=0-\dfrac{1}{9}\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}:3\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{27}\\ \left(3x-\dfrac{1}{2}\right)^3=\left(\dfrac{-1}{3}\right)^3\\ \Leftrightarrow3x-\dfrac{1}{2}=\dfrac{-1}{3}\\ 3x=\dfrac{-1}{3}+\dfrac{1}{2}\\ 3x=\dfrac{1}{6}\\ x=\dfrac{1}{6}:3\\ x=\dfrac{1}{18}\)

tìm x a)
\(\dfrac{7}{2}\)-\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{4}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{4}\) + \(\dfrac{7}{2}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{12}+\dfrac{7}{12}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-12}{12}=1\)
\(x+\dfrac{7}{10}\)= 1 . \(\dfrac{6}{5}\)
*Rồi tự làm phần tt đi

Mình ghi kết quả luôn nha bạn

c) \(\dfrac{x+1}{35}+\dfrac{x+2}{34}+\dfrac{x+3}{33}=\dfrac{x+4}{32}+\dfrac{x+5}{31}+\dfrac{x+6}{30}\)

\(\Rightarrow\dfrac{x+1}{35}+1+\dfrac{x+2}{34}+1+\dfrac{x+3}{33}+1=\dfrac{x+4}{32}+1+\dfrac{x+5}{31}+1+\dfrac{x+6}{30}+1\)

\(\Rightarrow\dfrac{x+1+35}{35}+\dfrac{x+2+34}{34}+\dfrac{x+3+33}{33}=\dfrac{x+4+32}{32}+\dfrac{x+5+31}{31}+\dfrac{x+6+30}{30}\)

\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}=\dfrac{x+36}{32}+\dfrac{x+36}{31}+\dfrac{x+36}{30}\)

\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}-\dfrac{x+36}{32}-\dfrac{x+36}{31}-\dfrac{x+36}{30}=0\)

\(\Rightarrow\left(x+36\right)\left(\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\right)=0\)

\(\Rightarrow x+36=0\left(\text{vì }\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\ne0\right)\)

\(\Rightarrow x=-36\)

Vậy ...

a/ Ta có: \(-4\dfrac{3}{5}.2\dfrac{4}{3}\le x\le-2\dfrac{3}{5}:1\dfrac{6}{15}\)

\(\Rightarrow\dfrac{-23}{5}.\dfrac{10}{3}\le x\le\dfrac{-13}{5}:\dfrac{21}{15}\)

\(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{5}.\dfrac{15}{21}\)

\(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{7}\)

\(\Rightarrow-15,\left(3\right)\le x\le-1,\left(857142\right)\)

Vì x \(\in\) Z nên x \(\in\left\{-1;-2;-3;...;-15\right\}\)

Chúc bạn học tốt!!!

a) \(\left(2x-3\right)\left(6-2x\right)=0\)

\(\circledast\)TH₁: \(2x-3=0\\ 2x=0+3\\ 2x=3\\ x=\dfrac{3}{2}\)

\(\circledast\)TH₂: \(6-2x=0\\ 2x=6-0\\ 2x=6\\ x=\dfrac{6}{2}=3\)

Vậy \(x\in\left\{\dfrac{3}{2};3\right\}\).

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

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

\(\dfrac{1}{3}x=-\dfrac{2}{5}\left(x-1\right)\)

\(-\dfrac{2}{5}-\dfrac{1}{3}=-x\left(x-1\right)\)

\(-\dfrac{11}{15}=-x\left(x-1\right)\)

\(\Rightarrow x=1.491631652\)

Vậy \(x=1.491631652\)

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

\(\circledast\)TH₁: \(3x-1=0\\ 3x=0+1\\ 3x=1\\ x=\dfrac{1}{3}\)

\(\circledast\)TH₂: \(-\dfrac{1}{2}x+5=0\\ -\dfrac{1}{2}x=0-5\\ -\dfrac{1}{2}x=-5\\ x=-5:-\dfrac{1}{2}\\ x=10\)

Vậy \(x\in\left\{\dfrac{1}{3};10\right\}\).

d) \(\dfrac{x}{5}=\dfrac{2}{3}\\ x=\dfrac{5\cdot2}{3}\\ x=\dfrac{10}{3}\)

Vậy \(x=\dfrac{10}{3}\).

e) \(\dfrac{x}{3}-\dfrac{1}{2}=\dfrac{1}{5}\\ \)

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

\(\dfrac{x}{3}=\dfrac{7}{10}\)

\(x=\dfrac{3\cdot7}{10}\)

\(x=\dfrac{21}{10}\)

Vậy \(x=\dfrac{21}{10}\).

f) \(\dfrac{x}{5}-\dfrac{1}{2}=\dfrac{6}{10}\)

\(\dfrac{x}{5}=\dfrac{6}{10}+\dfrac{1}{2}\)

\(\dfrac{x}{5}=\dfrac{11}{10}\)

\(x=\dfrac{5\cdot11}{10}\)

\(x=\dfrac{55}{10}=\dfrac{11}{2}\)

Vậy \(x=\dfrac{11}{2}\).

g) \(\dfrac{x+3}{15}=\dfrac{1}{3}\\ x+3=\dfrac{15}{3}=5\\ x=5-3\\ x=2\)

Vậy \(x=2\).

h) \(\dfrac{x-12}{4}=\dfrac{1}{2}\\ x-12=\dfrac{4}{2}=2\\ x=2+12\\ x=14\)

Vậy \(x=14\).

B=10 7/41-(2 7/41+5 3/4)

Giup minh nha!

vì mình ko thể đăng bài lên nên bạn thông cảm nha!

Bài 1: 

a: \(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2.15+3.75=\dfrac{8}{5}\)

=>x+4/15=8/5 hoặc x+4/15=-8/5

=>x=4/3 hoặc x=-28/15

b: \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{3}x=-\dfrac{1}{6}\\\dfrac{5}{3}x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{6}:\dfrac{5}{3}=\dfrac{-3}{30}=\dfrac{-1}{10}\\x=\dfrac{1}{10}\end{matrix}\right.\)

c: \(\Leftrightarrow\left|x-1\right|-1=1\)

=>|x-1|=2

=>x-1=2 hoặc x-1=-2

=>x=3 hoặc x=-1

Bài 2: 

b: \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{9}{25}\)

Bài 3: 

a: \(A=\left|x+\dfrac{15}{19}\right|-1>=-1\)

Dấu '=' xảy ra khi x=-15/19

b: \(\left|x-\dfrac{4}{7}\right|+\dfrac{1}{2}>=\dfrac{1}{2}\)

Dấu '=' xảy ra khi x=4/7

a) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)

\(-\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{1}{5}\)

\(-\dfrac{2}{3}x=\dfrac{1}{10}\)

x=\(\dfrac{1}{10}:-\dfrac{2}{3}\)

\(x=-\dfrac{3}{20}\)

Vậy \(x=-\dfrac{3}{20}\).

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

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

\(\dfrac{2}{3}:x=-\dfrac{22}{3}\)

\(x=\dfrac{2}{3}:-\dfrac{22}{3}\)

\(x=-\dfrac{1}{11}\)

Vậy \(x=-\dfrac{1}{11}\).

c) \(60\%x=\dfrac{1}{3}\cdot6\dfrac{1}{3}\)

\(60\%x=\dfrac{19}{9}\)

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

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

\(x=\dfrac{95}{27}\)

Vậy \(x=\dfrac{95}{27}\).

d) \(\left(\dfrac{2}{3}-x\right):\dfrac{3}{4}=\dfrac{1}{5}\)

\(\dfrac{2}{3}-x=\dfrac{1}{5}\cdot\dfrac{3}{4}\)

\(\dfrac{2}{3}-x=\dfrac{3}{20}\)

\(x=\dfrac{2}{3}-\dfrac{3}{20}\)

\(x=\dfrac{31}{60}\)

Vậy \(x=\dfrac{31}{60}\).

e) \(-2x-\dfrac{-3}{5}:\left(-0.5\right)^2=-1\dfrac{1}{4}\)

\(-2x-\dfrac{-12}{5}=-1\dfrac{1}{4}\)

\(-2x=-1\dfrac{1}{4}+\dfrac{-12}{5}\)

\(-2x=-\dfrac{73}{20}\)

\(x=-\dfrac{73}{20}:\left(-2\right)\)

\(x=\dfrac{73}{40}\)

Vậy \(x=\dfrac{73}{40}\).

..

K chép lại đề, lm luôn nhé:

*\(\Rightarrow\) \(\left(\dfrac{7}{2}+2x\right)\cdot\dfrac{8}{3}=\dfrac{16}{3}\)

\(\Rightarrow\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{8}{3}=2\)

\(\Rightarrow2x=2-\dfrac{7}{2}=-\dfrac{3}{2}\)

\(\Rightarrow x=-\dfrac{3}{4}\)

* \(\Rightarrow\left|2x-\dfrac{2}{3}\right|=\dfrac{\dfrac{3}{4}-2}{2}=-\dfrac{5}{8}\)

=> K có gt x nào t/m đề

* Đề sai

* \(\Rightarrow\left[{}\begin{matrix}3x-1=0\\-\dfrac{1}{2}x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)

*\(\Rightarrow\dfrac{1}{3}:\left(2x-1\right)=-5-\dfrac{1}{4}=-\dfrac{21}{4}\)

\(\Rightarrow2x-1=\dfrac{1}{3}:\left(-\dfrac{21}{4}\right)=-\dfrac{4}{63}\)

\(\Rightarrow2x=-\dfrac{4}{63}+1=\dfrac{59}{63}\)

\(\Rightarrow x=\dfrac{59}{63}:2=\dfrac{59}{126}\)

* \(\Rightarrow\left(2x+\dfrac{3}{5}\right)^2=\dfrac{9}{25}\)

\(\Rightarrow\left[{}\begin{matrix}2x+\dfrac{3}{5}=\dfrac{3}{5}\\2x+\dfrac{3}{5}=-\dfrac{3}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=0\Rightarrow x=0\\2x=-\dfrac{6}{5}\Rightarrow x=-\dfrac{3}{5}\end{matrix}\right.\)

* \(\Rightarrow-5x-1-\dfrac{1}{2}x+\dfrac{1}{3}=\dfrac{3}{2}x-\dfrac{5}{6}\)

\(\Rightarrow-5x-\dfrac{1}{2}x-\dfrac{3}{2}x=-\dfrac{5}{6}+1-\dfrac{1}{3}\)

\(\Rightarrow-7x=-\dfrac{1}{6}\)

\(\Rightarrow x=-\dfrac{1}{6}:7=-\dfrac{1}{42}\)

a)\(\left(3\dfrac{1}{2}+2x\right).2\dfrac{2}{3}=5\dfrac{1}{3}\)

\(\left(\dfrac{7}{2}+2x\right).\dfrac{8}{3}=\dfrac{16}{3}\)

\(\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{8}{3}=2\)

\(2x=2-\dfrac{7}{2}=\dfrac{-3}{2}\Rightarrow x=\dfrac{-3}{4}\)

b)\(\dfrac{3}{4}-2.\left|2x-\dfrac{2}{3}\right|=2\)

\(2.\left|2x-\dfrac{2}{3}\right|=\dfrac{3}{4}-2=\dfrac{-1}{4}\)

\(\Rightarrow\left|2x-3\right|=\dfrac{-1}{8}\)

\(\Rightarrow x\in\varnothing\)

c) Đề sai,bạn có viết chữ x đâu,đó là phép tính mà.

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

\(\Leftrightarrow3x-1=0\Rightarrow x=\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{-1}{2}x+5=0\Rightarrow x=10\)

e)\(\dfrac{1}{4}+\dfrac{1}{3}:\left(2x-1\right)=-5\)

\(\dfrac{1}{3}:\left(2x-1\right)=-5-\dfrac{1}{4}=\dfrac{-21}{4}\)

\(2x-1=\dfrac{1}{3}:\dfrac{-21}{4}=\dfrac{-4}{63}\)

\(\Rightarrow2x=\dfrac{59}{63}\Rightarrow x=\dfrac{59}{126}\)

g)\(\left(2x+\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\)

\(\left(2x+\dfrac{3}{5}\right)^2=0+\dfrac{9}{25}=\dfrac{9}{25}\)

\(\dfrac{9}{25}=\left(\dfrac{3}{5}\right)^2=\left(\dfrac{-3}{5}\right)^2\)

\(th1:x=0\)

\(th2:x=\dfrac{-3}{5}\)

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

\(-5x+-1-\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{3}{2}x-\dfrac{5}{6}\)

\(\Leftrightarrow-5x+-1+\dfrac{5}{6}-\dfrac{1}{3}=2x\)

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

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

\(\dfrac{-1}{2}=7x\Rightarrow x=\dfrac{-1}{14}\)

\(a.-8:\left(4\dfrac{1}{5}x+\dfrac{3}{10}\right)=4\dfrac{4}{9}\)

\(4\dfrac{1}{5}x+\dfrac{3}{10}=\left(-8\right):4\dfrac{4}{9}\)

\(4\dfrac{1}{5}x+\dfrac{3}{10}=\dfrac{-9}{5}\)

\(4\dfrac{1}{5}x=\dfrac{-9}{5}-\dfrac{3}{10}\)

\(4\dfrac{1}{5}x=\dfrac{-21}{10}\)

\(x=\dfrac{-21}{10}:\dfrac{21}{5}\)

\(x=\dfrac{-1}{2}\)

Vay \(x=\dfrac{-1}{2}\).

\(b.4\dfrac{2}{3}-\left(\dfrac{3}{5}:x\right)=-20\%\)

\(\dfrac{14}{3}-\left(\dfrac{3}{5}:x\right)=\dfrac{-1}{5}\)

\(\dfrac{3}{5}:x=\dfrac{14}{3}-\dfrac{-1}{5}\)

\(\dfrac{3}{5}:x=\dfrac{73}{15}\)

\(x=\dfrac{3}{5}:\dfrac{73}{15}\)

\(x=\dfrac{9}{73}\)

Vay \(x=\dfrac{9}{73}\).

Câu c; d; e tương tự nhé.