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

\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\2x-\frac{1}{3}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=1\\2x=\frac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{5}\\x=\frac{1}{6}\end{matrix}\right.\)

e, \(-\frac{3}{4}-\left|\frac{4}{5}-x\right|=-1\)

\(\Leftrightarrow\left|\frac{4}{5}-x\right|=-\frac{3}{4}-\left(-1\right)\)

\(\Leftrightarrow\left|\frac{4}{5}-x\right|=\frac{1}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{4}{5}-x=\frac{1}{4}\\\frac{4}{5}-x=-\frac{1}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{15}\\x=1,05\end{matrix}\right.\)

Vậy ....

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

\(\Rightarrow\left[\begin{matrix}5x-1=0\\2x-\frac{1}{3}=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}5x=1\\2x=\frac{1}{3}\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=\frac{1}{5}\\x=\frac{1}{6}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{1}{5};\frac{1}{6}\right\}\)

\(a)\)  Ta có : 

\(\frac{x}{18}=\frac{y}{9}\)\(\Leftrightarrow\)\(\frac{x}{2}=y\)

\(\Rightarrow\)\(x=2y\)

Thay \(x=2y\) vào \(A=\frac{2x-3y}{2x+3y}\) ta được : 

\(A=\frac{2.2y-3y}{2.2y+3y}=\frac{4y-3y}{4y+3y}=\frac{y}{7y}=\frac{1}{7}\)

Vậy ... ( tự kết luận ) 

Chúc bạn học tốt ~ 

ỳgyjwegfeukwfhưe

(5x-1) .(2x-1/3) =0

* 5x-1=0 * 2x-1/3=0

5x= o+1 2x=0+1/3

5x=1 2x=1/3

x=1/5 x=1/6

x=1/5; 1/6

(5x - 1).(2x - \(\frac{1}{3}\)) = 0

\(\left\{{}\begin{matrix}5x-1=0\\2x-\frac{1}{3}=0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}5x=0+1\\2x=0+\frac{1}{3}\end{matrix}\right.\) => \(\left\{{}\begin{matrix}5x=1\\2x=\frac{1}{3}\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=\frac{1}{5}\\x=\frac{1}{6}\end{matrix}\right.\)

Vậy x ∈ \(\left\{\frac{1}{5};\frac{1}{6}\right\}\)

Chúc bạn học tốt!

c)\(\left|2x+3\right|=x+2\)

Đk:\(x+2\ge0\Rightarrow x\ge-2\)

TH1:2x+3=x+2

\(\Rightarrow2x-x=2-3\)

\(\Rightarrow x=-1\)(Thỏa mãn đk )

TH2:2x+3=-x-2

\(\Rightarrow2x+x=-2+3\)

\(\Rightarrow3x=1\)

\(\Rightarrow x=\frac{1}{3}\)(Thỏa mãn đk)

Vậy x=-1 hoặc x=1/3

<==> ( 5x - 1)=0            < ===> 5x=1 <===>x=\(\frac{1}{5}\)

hoặc <===> (2x+\(\frac{1}{3}\)) =0        <===> 2x = \(\frac{-1}{3}\) <===> x =\(\frac{-1}{6}\)

vậy x = \(\frac{1}{5}\) hoặc x=\(\frac{-1}{6}\) nha bn !!!

bài này dễ ợt ah !!! cho mik đi :)))

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

\(\Leftrightarrow\hept{\begin{cases}5x-1=0\\2x+\frac{1}{3}=0\end{cases}\Leftrightarrow\hept{\begin{cases}5x=1\\2x=\frac{-1}{3}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{1}{5}\\x=\frac{-1}{6}\end{cases}}}\)

\(\text{Vậy x }\varepsilon\left(\frac{1}{5};\frac{-1}{6}\right)\)