\(1,\dfrac{4x-4}{3}=\dfrac{7-x}{5}\\ \Leftrightarrow5\left(4x-4\right)=3\left(7-x\right)\\ \Leftrightarrow20x-20=21-3x\\ \Leftrightarrow17x=41\Leftrightarrow x=\dfrac{41}{17}\)

\(2,\dfrac{3x-9}{5}=\dfrac{3-x}{2}\\ \Leftrightarrow6x-18=15-5x\\ \Leftrightarrow11x=33\\ \Leftrightarrow x=3\)

\(3,\dfrac{2x-1}{5}-\dfrac{3-x}{3}=1\\ \Leftrightarrow\dfrac{6x-3-15+5x}{15}=1\\ \Leftrightarrow11x-18=1\\ \Leftrightarrow x=\dfrac{19}{11}\)

\(4,\dfrac{x-5}{3}+\dfrac{3x+4}{2}=\dfrac{5x+2}{6}\\ \Leftrightarrow2x-10+9x+12=5x+2\\ \Leftrightarrow6x=0\Leftrightarrow x=0\)

\(5,\dfrac{x-3}{2}+\dfrac{2x+3}{5}=\dfrac{2x+5}{10}\\ \Leftrightarrow5x-15+4x+6=2x+5\\ \Leftrightarrow7x=14\\ \Leftrightarrow x=2\)

Tick nha

2: Ta có: \(\dfrac{3x-9}{5}=\dfrac{3-x}{2}\)

\(\Leftrightarrow6x-18=15-5x\)

\(\Leftrightarrow11x=33\)

hay x=3

\(\left(3-x\right)^3=-\dfrac{27}{64}\)

\(\left(3-x\right)^3=\left(\dfrac{-3}{4}\right)^3\)

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

\(x=3-\dfrac{-3}{4}=\dfrac{12}{4}+\dfrac{3}{4}\)

\(x=\dfrac{15}{4}\)

________

\(\left(x-5\right)^3=\dfrac{1}{-27}\)

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

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

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

\(x=\dfrac{14}{3}\)

_____________

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8}\)

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

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

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

\(x=2\)

________

\(\left(2x-1\right)^2=\dfrac{1}{4}\)            

\(\left(2x-1\right)^2=\left(\dfrac{1}{2}\right)^2\)           hoặc              \(\left(2x-1\right)^2=\left(\dfrac{-1}{2}\right)^2\)

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

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

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

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

\(x=\dfrac{3}{4}\)                                                       \(x=\dfrac{1}{4}\)

____________

\(\left(2-3x\right)^2=\dfrac{9}{4}\)

\(\left(2-3x\right)^2=\left(\dfrac{3}{2}\right)^2\)                hoặc                  \(\left(2-3x\right)^2=\left(\dfrac{-3}{2}\right)^2\)

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

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

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

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

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

______________

\(\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\) -> Kiểm tra đề câu này

(3-x)^{3_{=(-\(\dfrac{3}{4}\))}3}

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

  ^{x=3-(-\(\dfrac{3}{4}\))}

  ^{x=\(\dfrac{15}{4}\)}

1: Ta có: \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)

\(\Leftrightarrow5x+20+12x-28=7x+2\)

\(\Leftrightarrow17x-7x=2+8=10\)

hay x=1

2: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)

\(\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}\)

\(\Leftrightarrow6x+4-12x=-3x+3\)

\(\Leftrightarrow-6x+3x=3-4\)

hay \(x=\dfrac{1}{3}\)

3: Ta có: \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)

\(\Leftrightarrow4x-12-x-2=6x-3\)

\(\Leftrightarrow3x-14-6x+3=0\)

\(\Leftrightarrow-3x=11\)

hay \(x=-\dfrac{11}{3}\)

4: Ta có: \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)

\(\Leftrightarrow3x-6-8x-12=x+6\)

\(\Leftrightarrow-5x-x=6+18\)

hay x=-4

5: Ta có: \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)

\(\Leftrightarrow6x-3+2x-6=-1\)

\(\Leftrightarrow8x=8\)

hay x=1

câu E

\(\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left(2x-5\right)\left(5-2x\right)=-\left(\dfrac{3}{2}\right)^4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left|2x-5\right|=\left(\dfrac{3}{2}\right)^2\end{matrix}\right.\)

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\2x-5=-\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{11}{8}< \dfrac{5}{2}\left(n\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{5}{2}\\2x-5=\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{29}{8}>\dfrac{5}{2}\left(n\right)\end{matrix}\right.\end{matrix}\right.\)

câu F (bạn cho vào lớp 7.2=lớp 14 nhé. )

help 

-2/15

-4/35

-5/3

10

-8/3

-7/2

-9

-4/3

Chúc em học giỏi

\(=\dfrac{-2}{15}\\ =\dfrac{-4}{35}\\ =-1\\ =10\\ =\dfrac{-8}{3}\\ =-7\\ =-9\\ =\dfrac{-4}{3}\)

-2/15

-4/35

-5/3

10

-8/3

-7/2

-9

-4/3

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

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

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

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

\(\Leftrightarrow-x=-44\)

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

\(2,\left(đk:x\ne4\right)\)

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

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

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

\(\Leftrightarrow-x=13\)

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

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

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

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

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

a) \(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\\ < =>x\left(\dfrac{2}{3}-\dfrac{3}{2}\right)=\dfrac{5}{12}\\ < =>-\dfrac{5}{6}x=\dfrac{5}{12}\\ =>x=\dfrac{\dfrac{5}{12}}{-\dfrac{5}{6}}=-\dfrac{1}{2}\)

b) \(\dfrac{2}{5}+\dfrac{3}{5}.\left(3x-3,7\right)=\dfrac{-53}{10}\\ < =>\dfrac{2}{5}+\dfrac{9}{5}x-\dfrac{111}{50}=-\dfrac{53}{10}\\ < =>\dfrac{9}{5}x=-\dfrac{2}{5}+\dfrac{111}{50}-\dfrac{53}{10}\\ < =>\dfrac{9}{5}x=-\dfrac{87}{25}\\ =>x=\dfrac{\dfrac{-87}{25}}{\dfrac{9}{5}}=-\dfrac{29}{15}=-1\dfrac{14}{15}\)

c) \(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)+\dfrac{5}{9}=\dfrac{23}{27}\\ < =>\dfrac{\dfrac{7}{9}}{2}+\dfrac{\dfrac{7}{9}}{\dfrac{3}{4}x}+\dfrac{5}{9}=\dfrac{23}{27}\\ < =>\dfrac{7}{18}+\dfrac{28}{27}x+\dfrac{5}{9}=\dfrac{23}{27}\\ < =>\dfrac{28}{27}x=-\dfrac{7}{18}-\dfrac{5}{9}+\dfrac{23}{27}\\ < =>\dfrac{28}{27}x=\dfrac{1}{54}\\ =>x=\dfrac{\dfrac{1}{54}}{\dfrac{28}{27}}=\dfrac{1}{56}\)

d) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\\ < =>-\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{1}{5}\\ < =>-\dfrac{2}{3}x=\dfrac{1}{10}\\ =>x=\dfrac{\dfrac{1}{10}}{-\dfrac{2}{3}}=-\dfrac{3}{20}\)

e) \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\\ =>\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}\\ =>\left|x\right|=\dfrac{29}{12}\)

Vậy: x= 29/12 hoặc x= -29/12

f) \(\left|2x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\\ =>\left|2x-\dfrac{1}{3}\right|=1-\dfrac{5}{6}=\dfrac{1}{6}\\ =>\left[{}\begin{matrix}\left|2x-\dfrac{1}{3}\right|=\dfrac{1}{6}\\\left|2x-\dfrac{1}{3}\right|=-\dfrac{1}{6}\end{matrix}\right.\\ =>x=\left[{}\begin{matrix}x=\dfrac{\dfrac{1}{6}+\dfrac{1}{3}}{2}\\x=\dfrac{\dfrac{-1}{6}+\dfrac{1}{3}}{2}\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{1}{12}\end{matrix}\right.\)

Vậy: x= 1/4 hoặc x= 1/12