Đáp án:

`a,` `x∈{-1/16 ; -15/16}`

`b,` `(x,y)∈∅`

`----------`

`a,` `{|4x + 2|}/3- 3/4= -1/6`

`↔` `{|4x+2|}/3=-1/6+3/4`

`↔` `{|4x+2|}/3=7/12`

`↔` `|4x+2|=7/12 .3`

`↔` `|4x+2|=7/4`

`↔` \(\left[ \begin{array}{l}4x+2=\dfrac{7}{4}\\4x+2=-\dfrac{7}{4}\end{array} \right.\) 

`↔` \(\left[ \begin{array}{l}4x=\dfrac{7}{4}-2\\4x=-\dfrac{7}{4}-2\end{array} \right.\) 

`↔` \(\left[ \begin{array}{l}4x=-\dfrac{1}{4} \\4x=-\dfrac{15}{4}\end{array} \right.\) 

`↔` \(\left[ \begin{array}{l}x=-\dfrac{1}{4}:4\\x=-\dfrac{15}{4}:4\end{array} \right.\) 

`→` \(\left[ \begin{array}{l}x=-\dfrac{1}{16}\\x=-\dfrac{15}{16}\end{array} \right.\) 

Vậy `x∈{-1/16 ; -15/16}`

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`b,` `x/2-7/y=3/5`

`↔` `(5x)/10-70/(10y)=6/10`

`↔` $\begin{cases}5x-70=6\\10y=10\end{cases}$

`↔` $\begin{cases}5x=76\\10y=10\end{cases}$

`→` $\begin{cases}x=\dfrac{76}{5}\\y=1\end{cases}$

Vì `x,y∈NN^{**}` mà `x∉NN^(**);y∈NN^{**}` nên `(x,y)` không có giá trị thỏa mãn.

Vậy `(x,y)∈∅`

a. 

Ta có: |$\frac{4x+2}{3}$| - $\frac{3}{4}$ = $\frac{-1}{6}$ 

⇒ |$\frac{4x+2}{3}$| = $\frac{-1}{6}$ + $\frac{3}{4}$

⇒ |$\frac{4x+2}{3}$| = $\frac{7}{12}$ 

⇒ \(\left[ \begin{array}{l}\frac{4x+2}{3}=\frac{7}{12} \\\frac{4x+2}{3}=\frac{-7}{12}\end{array} \right.\) 

⇒ \(\left[ \begin{array}{l}\frac{16x+8}{12}=\frac{7}{12} \\\frac{16x+8}{12}=\frac{-7}{12}\end{array} \right.\) 

⇒ \(\left[ \begin{array}{l}16x+8=7\\16x+8=-7\end{array} \right.\) 

⇒ \(\left[ \begin{array}{l}16x=-1\\16x=-15\end{array} \right.\) 

⇒ \(\left[ \begin{array}{l}x=\frac{-1}{16} \\x=\frac{-15}{16} \end{array} \right.\) 

vậy x ∈ {$\frac{-1}{16}$;$\frac{-15}{16}$}

b. Ta có: $\frac{x}{2}$ - $\frac{7}{y}$ = $\frac{3}{5}$ 

⇒ $\frac{xy}{2y}$ - $\frac{14}{2y}$ = $\frac{3}{5}$ 

⇒ $\frac{xy-14}{2y}$ = $\frac{3}{5}$ 

⇒ $\frac{5xy-70}{10y}$ = $\frac{6y}{10y}$ 

⇒ 5xy-70 = 6y

⇒ 5xy - 6y = 70

⇒ y(5x-6) = 70

⇒ 70 chia hết cho y

⇒ y ∈ Ư(70)

⇒ y ∈ {1;2;5;7;10;14;35;70}

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