\(\left|x-1\right|=x+3\left(ĐK:x\ge-3\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-1=x+3\\x-1=-x-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-x=1+3\\x+x=1-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}0x=4\\2x=-2\end{matrix}\right.\\ \Leftrightarrow x=-1\left(tmđk\right)\)

Vậy x = -1 là nghiệm của pt.

Ta có : \(\dfrac{x}{2}=\dfrac{1-x}{3}\)

\(\Leftrightarrow3x=2\left(1-x\right)\)

\(\Leftrightarrow3x=2-2x\)

\(\Leftrightarrow5x=2\)

\(\Leftrightarrow x=\dfrac{5}{2}\)

Vậy ...

Ta có: \(\dfrac{x}{2}=\dfrac{1-x}{3}\)

\(\Leftrightarrow3x=2\left(1-x\right)\)

\(\Leftrightarrow3x=2-2x\)

\(\Leftrightarrow3x+2x=2\)

\(\Leftrightarrow5x=2\)

hay \(x=\dfrac{2}{5}\)

Vậy: \(x=\dfrac{2}{5}\)

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

D min khi x-3=-1

=>x=2

=>x=7/3:2/3=7/2

X là : 21/6

\(x\times\dfrac{2}{3}=\dfrac{7}{3}\)

\(x=\dfrac{7}{3}:\dfrac{2}{3}\)

\(x=\dfrac{7}{2}\)

\(\Leftrightarrow\left(x^2+2x\right)^2-2\left(x^2+2x\right)-3=0\)

\(\Leftrightarrow\left(x^2+2x-3\right)\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-1\right)\left(x+1\right)^2=0\)

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

     30% . x - x - \(\dfrac{5}{6}\) = \(\dfrac{1}{3}\)

\(\Rightarrow\) (\(\dfrac{3}{10}\) - 1) . x = \(\dfrac{1}{3}+\dfrac{5}{6}\)

\(\Rightarrow\) \(-\dfrac{7}{10}\) . x = \(\dfrac{7}{6}\)

\(\Rightarrow\) x = \(-\dfrac{3}{5}\)

KO ghi đề nhé

\(\dfrac{3}{10}.x-x=\dfrac{1}{3}+\dfrac{5}{6}\)

\(x\left(\dfrac{3}{10}-1\right)=\dfrac{2}{6}+\dfrac{5}{6}\)

\(x.\dfrac{-7}{10}=\dfrac{7}{6}\)

\(x=\dfrac{7}{10}:\dfrac{-7}{6}\)

\(x=\dfrac{7}{10}.\dfrac{6}{-7}\)

\(x=\dfrac{42}{-70}\)

\(x=\dfrac{6}{-10}\)

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

Vậy \(x=\dfrac{3}{-5}\)

\(30\%.x-x-\dfrac{5}{6}=\dfrac{1}{3}\) 

     \(x.\left(30\%-1\right)=\dfrac{1}{3}+\dfrac{5}{6}\) 

               \(x.\dfrac{-7}{10}=\dfrac{7}{6}\)  

                       \(x=\dfrac{7}{6}:\dfrac{-7}{10}\) 

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

\(\Leftrightarrow2.5:x=2+\dfrac{2}{3}-2-\dfrac{1}{5}=\dfrac{7}{15}\)

hay \(x=\dfrac{5}{2}:\dfrac{7}{15}=\dfrac{5}{2}\cdot\dfrac{15}{7}=\dfrac{75}{14}\)

\(2^x+2^{x+1}+2^{x+2}=960-2^{x+3}\)

\(\Leftrightarrow2^x+2^{x+1}+2^{x+2}+2^{x+3}=960\)

\(\Leftrightarrow2^x\left(1+2+2^2+2^3\right)=960\)

\(\Leftrightarrow2^x.15=960\)

\(\Leftrightarrow2^x=64\)

\(\Leftrightarrow2^x=2^6\Leftrightarrow x=6\)

Vậy...