(\(x\) + \(\dfrac{1}{2}\))² = \(\dfrac{1}{16}\)

\(\left[{}\begin{matrix}x+\dfrac{1}{2}=-\dfrac{1}{4}\\x+\dfrac{1}{2}=\dfrac{1}{4}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=-\dfrac{1}{4}-\dfrac{1}{2}\\x=\dfrac{1}{4}-\dfrac{1}{2}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

Vậy \(x\) \(\in\) {- \(\dfrac{3}{4};-\dfrac{1}{4}\)}

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

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

\(x\)             = \(\dfrac{1}{81}\)

Vậy \(x=\dfrac{1}{81}\)

\(\left|x+1\right|+\left|6-x\right|=0\)

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

A= 2(1+2+ .. + 8)=2.55=110

=> A : 3 = 110 /3 = 36 R = 2

;

\(\frac{x}{4}=\frac{y}{5}\Rightarrow\frac{x}{8}=\frac{y}{10}\) (1)

\(\frac{y}{2}=\frac{z}{3}\Rightarrow\frac{y}{10}=\frac{z}{15}\)(2)

Từ (1) và (2) \(\Rightarrow\frac{x}{8}=\frac{y}{10}=\frac{z}{15}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta được : 

\(\frac{x}{8}=\frac{y}{10}=\frac{z}{15}=\frac{x-y-z}{8-10-15}=\frac{37}{-17}\)

\(\Rightarrow\frac{x}{8}=\frac{37}{-17}\Rightarrow x=-\frac{296}{17}\)

\(\Rightarrow\frac{y}{10}=-\frac{37}{17}\Rightarrow y=-\frac{370}{17}\)

\(\Rightarrow\frac{z}{15}=-\frac{37}{17}\Rightarrow z=-\frac{555}{17}\)