\(x+\dfrac{1}{2}=\dfrac{33}{4}\\ \Rightarrow x=\dfrac{33}{4}-\dfrac{1}{2}\\ \Rightarrow x=\dfrac{31}{4}\\ \dfrac{5}{6}-x=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{5}{6}-\dfrac{1}{3}\\ \Rightarrow x=\dfrac{1}{2}\\ x+\dfrac{4}{5}=\dfrac{-2}{3}\\ \Rightarrow x=\dfrac{-2}{3}-\dfrac{4}{5}\\ \Rightarrow x=\dfrac{-22}{15}\)

`(50% . x+5 1/4) . (-2/3) = 2 5/6`

`(1/2 x + 21/4) . (-2/3) = 17/6`

`1/2 x + 21/4 = -17/4`

`1/2 x = -19/2`

`x=-19`

\(^{^{tham}}\) \(^{^{khảo}}\)

\(\left(50\%.+5\dfrac{1}{4}\right).\left(-\dfrac{2}{3}\right)=2\dfrac{5}{6}\)

\(\left(\dfrac{1}{2}x+\dfrac{21}{2}\right).\left(-\dfrac{2}{3}\right)=\dfrac{17}{6}\)
\(\dfrac{1}{2}x+\dfrac{21}{4}=-\dfrac{17}{4}\)

\(\dfrac{1}{2}x=-\dfrac{19}{2}\)

\(x=-19\)

\(\left(50\%.x+5\dfrac{1}{4}\right).\dfrac{-2}{3}=2\dfrac{5}{6}\) 

                 \(\dfrac{1}{2}.x+\dfrac{21}{4}=\dfrac{17}{6}:\dfrac{-2}{3}\) 

                 \(\dfrac{1}{2}.x+\dfrac{21}{4}=\dfrac{-17}{4}\) 

                           \(\dfrac{1}{2}.x=\dfrac{-17}{4}-\dfrac{21}{4}\) 

                           \(\dfrac{1}{2}.x=\dfrac{-19}{2}\) 

                                \(x=\dfrac{-19}{2}:\dfrac{1}{2}\) 

                                \(x=-19\)

\(x-\dfrac{3}{5}=\dfrac{2}{7}+\dfrac{1}{6}\)

\(\Leftrightarrow x-\dfrac{3}{5}=\dfrac{19}{42}\)

\(\Leftrightarrow x=\dfrac{19}{42}+\dfrac{3}{5}\)

\(\Leftrightarrow x=\dfrac{221}{210}\)

=>x-3/5=12/42+7/42=19/42

=>x=19/42+3/5=95/210+126/210=221/210

x-3/5=2/7+1/6

x-3/5=12/42+7/42

x-3/5=19/42

x=19/42+3/5

x=95/210 + 126/210

x=221/210

\(TH1:x\ge0\)

\(\Rightarrow x\left(1-\dfrac{5}{6}\right)=\dfrac{4}{9}.\dfrac{15}{8}\)

\(\Rightarrow x=\dfrac{\dfrac{4}{9}.\dfrac{15}{8}}{1-\dfrac{5}{6}}=5\left(TM\right)\)

\(TH2:x< 0\)

\(\Rightarrow x\left(-1-\dfrac{5}{6}\right)=\dfrac{4}{9}.\dfrac{15}{8}\)

\(\Rightarrow x=\dfrac{\dfrac{4}{9}.\dfrac{15}{8}}{-1-\dfrac{5}{6}}=-\dfrac{5}{11}\left(TM\right)\)

Vậy ...

Giải:

\(\left|x\right|-\dfrac{5}{6}.x=\dfrac{4}{9}.\dfrac{15}{8}\) 

\(TH1:x\ge0\) 

\(\left|x\right|-\dfrac{5}{6}.x=\dfrac{4}{9}.\dfrac{15}{8}\) 

  \(x-\dfrac{5}{6}.x=\dfrac{5}{6}\)   

\(x.\left(1-\dfrac{5}{6}\right)=\dfrac{5}{6}\) 

         \(x.\dfrac{1}{6}=\dfrac{5}{6}\) 

              \(x=\dfrac{5}{6}:\dfrac{1}{6}\) 

              \(x=5\) 

\(TH2:x\le0\) 

\(\left|x\right|-\dfrac{5}{6}.x=\dfrac{4}{9}.\dfrac{15}{8}\) 

\(-x-\dfrac{5}{6}.x=\dfrac{5}{6}\) 

\(x.\left(-1-\dfrac{5}{6}\right)=\dfrac{5}{6}\) 

     \(x.\dfrac{-11}{6}=\dfrac{5}{6}\) 

               \(x=\dfrac{5}{6}:\dfrac{-11}{6}\) 

               \(x=\dfrac{-5}{11}\) 

Vậy \(x\in\left\{\dfrac{-5}{11};5\right\}\)

(\(x\)-\(\dfrac{2}{3}\))^{2\(=\)\(\dfrac{5}{6}\)}

^\(x\)-\(\dfrac{4}{9}\)\(=\)\(\dfrac{5}{6}\)

\(x\)\(=\)\(\dfrac{5}{6}\)+\(\dfrac{4}{9}\)

\(x\)\(=\)\(\dfrac{23}{18}\)

\(\left(x-\dfrac{2}{3}\right)^2=\dfrac{5}{6}\)

\(< =>\left(x-\dfrac{2}{3}\right)^2-\dfrac{5}{6}=0\)

<=>\(\left(x-\dfrac{2}{3}\right)^2-\left(\dfrac{\sqrt{5}}{\sqrt{6}}\right)^2=0\)

\(< =>\left(x-\dfrac{2}{3}+\dfrac{\sqrt{5}}{\sqrt{6}}\right)\left(x-\dfrac{2}{3}-\dfrac{\sqrt{5}}{\sqrt{6}}\right)=0\)

=>\(\left[{}\begin{matrix}x-\dfrac{2}{3}+\dfrac{\sqrt{5}}{\sqrt{6}}=0\\x-\dfrac{2}{3}-\dfrac{\sqrt{5}}{\sqrt{6}}=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=\dfrac{2}{3}-\dfrac{\sqrt{5}}{\sqrt{6}}\\x=\dfrac{2}{3}+\dfrac{\sqrt{5}}{\sqrt{6}}\end{matrix}\right.\)

<=>\(\left[{}\begin{matrix}x=\dfrac{4-\sqrt{30}}{6}\\x=\dfrac{4+\sqrt{30}}{6}\end{matrix}\right.\)

VẬy..

x = 6; y=2

Lời giải:

$\frac{5}{x}-\frac{y}{3}=\frac{1}{6}$

$\Rightarrow \frac{15-xy}{3x}=\frac{1}{6}$

$\Rightarrow \frac{2(15-xy)}{6x}=\frac{x}{6x}$

$\Rightarrow 2(15-xy)=x$

$\Rightarrow 30=2xy+x$

$\Rightarrow 30=x(2y+1)$

$\Rightarrow x=\frac{30}{2y+1}$

Vì $x$ nguyên nên $\frac{30}{2y+1}$ nguyên

$\Rightarrow 2y+1$ là ước của $30$

Vì $2y+1$ lẻ nên $2y+1\in\left\{\pm 1; \pm 3; \pm 5; \pm 15\right\}$

$\Rightarrow y\in\left\{-1; 0; -2; 1; -3; 2; -8; 7\right\}$

Tương ứng với các giá trị $y$ trên ta có: $x\in\left\{-30; 30; -10; 10; -6; 6; -2;2\right\}$

\(\Rightarrow x-1=\dfrac{6}{7}+\dfrac{1}{7}\times\left(\dfrac{2}{7}+\dfrac{5}{7}\right)=\dfrac{6}{7}+\dfrac{1}{7}\times1\\ \Rightarrow x-1=\dfrac{6}{7}+\dfrac{1}{7}=1\\ \Rightarrow x=1+1=2\)

= 2

`x-(5/6-x)=x-2/3`

⇒`x-5/6+x=x-2/3`

⇒`2x-5/6=x-2/3`

⇒`2x-5/6-x+2/3=0`

⇒`x-1/6=0`

⇒`x=1/6`