x=5;y=2

a) x=4

\(\left(x-4\right)\left(x-6\right)< 0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-4< 0\\x-6>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-4>0\\x-6< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 4\\x>6\left(vô.lí\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x>4\\x< 6\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow4< x< 6\)

Mà x là số nguyên nên x=5

Để \(\left(x-4\right)\left(x-6\right)< 0\) thì \(\left(x-4\right)\) và \(\left(x-6\right)\) trái dấu

Hay \(x-4>0;x-6< 0\) => \(x>4;x< 6\Rightarrow x=\left\{5\right\}\)

hoặc \(x-4< 0;x-6>0\) => \(x< 4;x>6\Rightarrow x\in\varnothing\)

Vậy: Giá trị x nguyên thỏa mãn điều kiện là 5.

\(x+4⋮2x+1\)

=>\(2x+8⋮2x+1\)

=>\(2x+1+7⋮2x+1\)

=>\(7⋮2x+1\)

=>\(2x+1\in\left\{1;-1;7;-7\right\}\)

=>\(2x\in\left\{0;-2;6;-8\right\}\)

=>\(x\in\left\{0;-1;3;-4\right\}\)

Ta có:

(x + 4) ⋮ (2x + 1)

⇒ 2(x + 4) ⋮ (2x + 1)

⇒ (2x + 8) ⋮ (2x + 1)

⇒ (2x + 1 + 7) ⋮ (2x +1)

⇒ 7 ⋮ (2x + 1)

⇒ 2x + 1 ∈ Ư(7) = {-7; -1; 1; 7}

⇒ 2x ∈ {-8; -2; 0; 6}

⇒ x ∈ {-4; -1; 0; 3}

2x - 137 = 3²

2x - 137 = 9

2x = 9 + 137

2x = 146

x = 146 : 2

x = 73

\(2x-137=3^2\)

\(2x-137=9\)

\(2x=9+137\)

\(2x=146\)

\(x=146:2\)

\(x=73\)

học tốt

\(2x-137=3^2\)

\(2x-137=9\)

\(2x=9+137\)

\(2x=146\)

\(x=146\div2\)

\(x=73\)

Vậy x = 73

\(\frac{-6}{2x-4}\) nguyên

\(\Leftrightarrow-6⋮2x-4\)

\(\Rightarrow2x-4\inƯ\left(-6\right)\)

\(\Rightarrow2x-4\in\left\{-1;-2;-3;-6;1;2;3;6\right\}\)

\(\Rightarrow2x\in\left\{3;2;1;-2;5;6;7;10\right\}\)

\(\Rightarrow x\in\left\{1,5;1;0,5;-1;2,5;3;3,5;5\right\}\) mà \(x\in Z\)

\(\Rightarrow x\in\left\{1;-1;3;5\right\}\)

Chị rút gọn luôn đi cho số nó nhỏ hơn !!!

\(\left(x+4\right)⋮\left(2x+1\right)\Rightarrow2\left(x+4\right)=2x+8=2x+1+7⋮\left(2x+1\right)\Leftrightarrow7⋮\left(2x+1\right)\)

\(\Leftrightarrow2x+1\inƯ\left(7\right)=\left\{-7,-1,1,7\right\}\Leftrightarrow x\in\left\{-4,-1,0,3\right\}\).

Thử lại đều thỏa mãn.