-5 x ( x + 1/5 ) - 1/2 x ( x - 2/3 ) = 3/2x - 5/6

<=> -5x^2 - x - 1/2 x^2 + 1/3 = 3/2x - 5/6

<=> -3x^2 - x + 1/3 = 3/2x - 5/6

<=> -3x^2  - 5/2 x + 7/6 = 0

<=> ( 3x - 1 ) ( 6x + 7 ) = 0

=> x = 1/3 ; x = -7/6 .

Vậy x = 1/3 , x = -7/6.

\(\left(\frac{14}{5}x-50\right)\div\frac{2}{3}=51\)

\(\Leftrightarrow\frac{14}{5}x-50=51\times\frac{2}{3}\)

\(\Leftrightarrow\frac{14}{5}x-50=34\)

\(\Leftrightarrow\frac{14}{5}x=34+50\)

\(\Leftrightarrow\frac{14}{5}x=84\)

\(\Leftrightarrow x=84\div\frac{14}{5}\)

\(\Leftrightarrow x=84\times\frac{5}{14}\)

\(\Leftrightarrow x=30\)

\(\left(\frac{14}{5}.x-50\right):\frac{2}{3}=51\)

\(\frac{14}{5}.x-50=51.\frac{2}{3}\)

\(\frac{14}{5}.x-50=34\)

\(\frac{14}{5}.x=34+50\)

\(\frac{14}{5}.x=84\)

\(x=84:\frac{14}{5}\)

\(x=30\)

( 14/5 . x - 50 ) : 2/3 = 51

14/5 . x - 50 = 51 . 2/3

14/5 .x - 50 = 34

14/5.x = 34 + 50

14/5.x = 84

=> x = 30

(2,5x - 4/3)² - 5/6 = -7/18

=> (2,5x - 4/3)² = -7/18 + 5/6

=> (2,5x - 4/3)² = 4/9

=> (2,5x - 4/3)² = (2/3)²

=> \(\orbr{\begin{cases}2,5x-\frac{4}{3}=\frac{2}{3}\\2,5x-\frac{4}{3}=-\frac{2}{3}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{4}{5}\\x=\frac{4}{15}\end{cases}}\)

\(\left(2,5x-\frac{4}{3}\right)^2-\frac{5}{6}=-\frac{7}{18}\)

\(\left(2,5x-\frac{4}{3}\right)^2=-\frac{7}{18}+\frac{5}{6}\)

\(\left(2,5x-\frac{4}{3}\right)^2=\frac{4}{9}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{4}{5}\\x=\frac{4}{15}\end{cases}}\)

\(\left(\frac{3}{5}-\frac{2}{3}x\right)^3=\left(\frac{-4}{5}\right)^3\)

\(\frac{3}{5}-\frac{2}{3}x=\frac{-4}{5}\)

\(\frac{2}{3}x=\frac{7}{5}\)

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

\(\left(\frac{3}{5}-\frac{2}{3}.x\right)^3=\frac{-64}{125}\)

\(\Leftrightarrow\left(\frac{3}{5}-\frac{2}{3}.x\right)^3=\left(-\frac{4}{5}\right)^3\)

\(\Rightarrow\frac{3}{5}-\frac{2}{3}.x=-\frac{4}{5}\)

\(\Leftrightarrow\frac{2}{3}.x=\frac{3}{5}-\left(-\frac{4}{5}\right)\)

\(\Leftrightarrow\frac{2}{3}.x=\frac{3}{5}+\frac{4}{5}\)

\(\Leftrightarrow\frac{2}{3}.x=\frac{7}{5}\)

\(\Leftrightarrow x=\frac{7}{5}:\frac{2}{3}\)

\(\Leftrightarrow x=\frac{21}{10}\)

Vậy \(x=\frac{21}{10}\)

\(\left(\frac{3}{5}-\frac{2}{3}x\right)^3=\frac{-64}{125}\)

\(\left(\frac{3}{5}-\frac{2}{3}x\right)^3=\left(\frac{-4}{5}\right)^3\)

\(\frac{3}{5}-\frac{2}{3}x=\frac{-4}{5}\)

\(\frac{9}{15}-\frac{10x}{15}=\frac{-12}{15}\)

\(-\frac{10x}{15}=\frac{9}{15}-\frac{-12}{15}\)
\(\text{ }-10x=9+12\)

\(-10x=21\)

\(x=\frac{-21}{10}\)

k nha

ko ghi lại đề nha 

a ) x - 2 = 12 

    x       = 14 

b ) 5x + x = 39-3¹¹ : 3⁹

          6x  = 39 - 3²

           6x = 39 - 9

              6x= 30

             x   = 5 

c ) 3^x = 9 

     ^{x     = 2}   ( vì 3² = 9 ) 

^{d ) 4x = 64} 

       ^{x = 64/4 = 16} 

^{e )} 2^x = 16 

^{x       = 4 (} vì 2⁴ = 16 )

^{f )} 9^x-1 = 9 

^{x - 1      = 1 (} vì 9¹ = 9 ) 

^{x            =  2}

^{g )} x⁴ = 16

 ^{x        = 2} 

^{h )} 2^x : 2⁵ = 1 

   2^x-5 = 1 

^{x - 5     = 0 (} vì bất cứ số nào luỹ thừa 0 cũng bằng 1 ) 

^{x          = 5}   

(đã xóa câu trả lời) _by tth

\(x+y=5\)

,=>  \(\left(x+y\right)^3=125\)

<=>  \(x^3+y^3+3xy\left(x+y\right)=125\)

<=>  \(x^3+y^3+3.3.5=125\)

<=> \(x^3+y^3=80\)

Vậy...