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

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

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

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

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

________

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

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

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

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

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

_____________

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

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

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

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

\(x=2\)

________

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

\(\left(2x-1\right)^2=\left(\dfrac{1}{2}\right)^2\)           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=\(\dfrac{15}{4}\)}

đấy nhá

b) Ta có: \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)

\(\Leftrightarrow\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}=\dfrac{20}{12}+\dfrac{9}{12}=\dfrac{29}{12}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{12}\\x=-\dfrac{29}{12}\end{matrix}\right.\)

c) Ta có: \(\left|2x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\)

\(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{1}{6}\)

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

h) \(5^x+5^{x+2}=650\)

\(\Leftrightarrow5^x+5^x.5^2=650\)

\(\Leftrightarrow5^x\left(1+25\right)=650\)

\(\Leftrightarrow5^x.26=650\)

\(\Leftrightarrow5^x=25\)

\(\Leftrightarrow x=2\)

haizzz,đăng ít thôi,chứ nhìn hoa mắt quá =.=

-1 5/27-(3x-7/9)³=-24/27

           (3x-7/9)³ =-32/27-(-24/27)

           (3x-7/9)³=-8/27

           (3x-7/9)³=(-2/3)³

⇒3x-7/9=-2/3

   3x      =-2/3+7/9

   3x      =1/9

     x      =1/9:3

     x      =1/27

Chúc bạn học tốt!

a) Ta có: \(\dfrac{5}{8}+\dfrac{3}{17}+\dfrac{4}{18}+\dfrac{20}{-17}+\dfrac{-2}{9}+\dfrac{21}{56}\)

\(=\left(\dfrac{3}{17}-\dfrac{20}{17}\right)+\left(\dfrac{2}{9}-\dfrac{2}{9}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)\)

\(=-1+1=0\)

b) Ta có: \(\left(\dfrac{9}{16}+\dfrac{8}{-27}\right)+\left(1+\dfrac{7}{16}+\dfrac{-19}{27}\right)\)

\(=\left(\dfrac{9}{16}+\dfrac{7}{16}\right)+\left(\dfrac{-8}{27}-\dfrac{19}{27}\right)+1\)

=1-1+1=1

b: =>(3x-1)(3x+1)(2x+3)=0

hay \(x\in\left\{\dfrac{1}{3};-\dfrac{1}{3};-\dfrac{3}{2}\right\}\)

c: \(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{5}{6}+\dfrac{3}{4}=\dfrac{19}{12}\)

=>2x-1/3=19/12 hoặc 2x-1/3=-19/12

=>2x=23/12 hoặc 2x=-15/12=-5/4

=>x=23/24 hoặc x=-5/8

d: \(\Leftrightarrow-\dfrac{5}{6}\cdot x+\dfrac{3}{4}=-\dfrac{3}{4}\)

=>-5/6x=-3/2

=>x=3/2:5/6=3/2*6/5=18/10=9/5

e: =>2/5x-1/2=3/4 hoặc 2/5x-1/2=-3/4

=>2/5x=5/4 hoặc 2/5x=-1/4

=>x=5/4:2/5=25/8 hoặc x=-1/4:2/5=-1/4*5/2=-5/8

f: =>14x-21=9x+6

=>5x=27

=>x=27/5

h: =>(2/3)^2x+1=(2/3)^27

=>2x+1=27

=>x=13

i: =>5^3x*(2+5^2)=3375

=>5^3x=125

=>3x=3

=>x=1

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

a, Theo đề ta có:

\(2.3^x-405=3^{x-1}\)

=> \(2.3^x-405=3^x:3\)

=> \(405=(2.3^x)-(3^x:3)\)

=>\(405=(2.3^x)-(3^x.\dfrac{1}{3})\)

=> \(405=3^x(2-\dfrac{1}{3})\)

=>\(405=3^x(\dfrac{6}{3}-\dfrac{1}{3})\)

=> \(405=3^x.\dfrac{5}{3}\)

=> \(3^x=405:\dfrac{5}{3}\)

=>\(3^x=405.\dfrac{3}{5}\)

=> \(3^x=81.3\)

=> \(3^x=243\)

=> \(3^x=3^5\)

=> x=5

Vậy:..............................

a. x=5/6

\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)

\(a,\dfrac{2}{3}+\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}\)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}-\dfrac{2}{3}\)

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

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

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

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

\(x=\dfrac{1}{5}\)