\(\left(3x-2\right)^2=9\)

\(\left(3x-2\right)^2=\pm3^2\)

\(\Rightarrow\hept{\begin{cases}3x-2=3\\3x-2=\left(-3\right)\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\x=\left(-\frac{1}{3}\right)\end{cases}}\)

Vậy ...

ta có 3x = 4y 

=> x/4 = y/3

áp dụng tc của dãy tỉ số bằng nhau:

=> x/4 = y/3 = y-x / 3-4 = 5/-1 = -5

=>x/4 = -5 => x = 4 x -5 = -20

    y/3 = -5 => y = 3 x -5 = -15

vậy x, y = -20 , -15

HT

\(i,-\frac{7}{5}.\frac{3}{8}+\left(-\frac{7}{8}\right).\frac{13}{5}+\frac{3}{8}:\left(-\frac{6}{5}\right)\)

\(=-\frac{21}{40}+\left(-\frac{91}{40}\right)+\frac{3}{8}.\left(-\frac{5}{6}\right)\)

\(=-\frac{21}{40}-\frac{91}{40}-\frac{5}{16}\)

\(=-\frac{42}{80}-\frac{182}{80}-\frac{40}{80}\)

\(=-\frac{264}{80}\)

\(=\left(-\frac{33}{10}\right)\)

Thanks các bn nha

giup minh voi

\(\frac{3-x}{-2}\)= \(\frac{1}{-3}\) => (3-x) . (-3) =1 . (-2)    =>(3-x) . (-3) = -2    =>3-x = -2 : (-3)    => 3-x =\(\frac{2}{3}\)   => x=\(2\frac{1}{3}\)

\(\frac{3-x}{-2}=\frac{1}{-3}\)

\(\Leftrightarrow-3\left(3-x\right)=-2.1\)

\(-9+3x=-2\)

\(-9+3x+2=0\)

\(-7+3x=0\)

\(3x=7\)

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

\(\left(\frac{2}{3}\right)^{12}:\left(\frac{2}{3}\right)^{10}\)

\(=\left(\frac{2}{3}\right)^{12-10}\)

\(=\left(\frac{2}{3}\right)^2\)

\(=\frac{4}{9}\)

Vâyj ... 

kết bạn tui giải cho

\(\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\\frac{y}{5}=\frac{z}{4}\end{cases}}\Leftrightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có: 

\(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x-y+z}{10-15+12}=\frac{-49}{7}=-7\)

\(\Leftrightarrow\hept{\begin{cases}x=-7.10=-70\\y=-7.15=-105\\z=-7.12=-84\end{cases}}\)