1) 6x\(^2\) + 5x - 11 = 0

<=> 6x\(^2\) - 6x + 11x - 11 = 0

<=> 6x . (x - 1) + 11 . (x - 1) = 0

<=> (x - 1)(6x + 11) = 0

<=> \(\orbr{\begin{cases}x-1=0\\6x+11=0\end{cases}}\) <=> \(\orbr{\begin{cases}x=1\\6x=-11\end{cases}}\) <=> \(\orbr{\begin{cases}x=1\\x=-\frac{11}{6}\end{cases}}\)

2) 7x\(^2\) - 4x - 3 = 0

<=> 7x\(^2\) - 7x + 3x - 3 = 0

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

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

<=> \(\orbr{\begin{cases}x-1=0\\7x+3=0\end{cases}}\) <=> \(\orbr{\begin{cases}x=1\\7x=-3\end{cases}}\) <=> \(\orbr{\begin{cases}x=1\\x=-\frac{3}{7}\end{cases}}\)

3) 5x\(^2\) - 2x - 3 = 0

<=> 5x\(^2\) - 5x + 3x - 3 = 0

<=> 5x . (x - 1) + 3 . (x - 1) = 0

<=> (x - 1)(5x + 3) = 0

<=> \(\orbr{\begin{cases}x-1=0\\5x+3=0\end{cases}}\) <=> \(\orbr{\begin{cases}x=1\\5x=-3\end{cases}}\) <=> \(\orbr{\begin{cases}x=1\\x=-\frac{3}{5}\end{cases}}\)

(5x-1)(2x- \(\frac{1}{3}\)) = 0

=> \(\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}}\)=> \(\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}}\)=> \(\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)

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

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

Vậy \(x=\dfrac{1}{5}\) hoặc \(x=\dfrac{1}{6}\)

Mình ko chắc là mình làm đúng 100 % đâu nên nếu mình làm sai thì bạn thông cảm nhé !

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

\(\Rightarrow\) 5x - 1 = 0 hoặc 2x - \(\dfrac{1}{3}\) = 0

5x = 0+1 2x = 0 + \(\dfrac{1}{3}\)

5x = 1 2x = \(\dfrac{1}{3}\)

x = \(\dfrac{1}{5}\) x = \(\dfrac{1}{3}\) \(_{\div}\) 2 = \(\dfrac{1}{3}\times2\) = \(\dfrac{2}{3}\)

Vậy x= \(\dfrac{1}{5}\) hoặc x= \(\dfrac{2}{3}\)

Chúc bạn học tốt !

(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!

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 ....

<==> ( 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 :)))

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

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

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

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

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

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

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

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

Vậy ...

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

\(\Rightarrow\) TH1 : 5x-1=0 \(\Rightarrow\) 5x=1 \(\Rightarrow\) x=\(\dfrac{1}{5}\)

\(\Rightarrow\) TH2 : 2x-\(\dfrac{1}{3}\) = 0 \(\Rightarrow\) 2x=\(\dfrac{1}{3}\) \(\Rightarrow\) x=\(\dfrac{1}{6}\)

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