a) \(\dfrac{x}{-6}=\dfrac{-15}{45}\)

    \(\dfrac{-x}{6}=\dfrac{-15}{45}\)

    \(\dfrac{x}{6}=\dfrac{15}{45}\)

    \(x=\dfrac{\left(15\cdot6\right)}{45}\)

    \(x=2\)

b) \(\dfrac{x}{5}=\dfrac{16}{25}\)

    \(x=\dfrac{\left(16\cdot5\right)}{25}\)

    \(x=\dfrac{80}{25}\)

   \(x=\dfrac{16}{5}\)

c) \(\dfrac{5}{x-3}=\dfrac{20}{-12}\)

  \(x-3=\dfrac{\left(5\cdot-12\right)}{20}\)

 \(x-3=-3\)

 \(x=\left(-3\right)+3\)

\(x=0\)

d) \(\dfrac{2}{5}\cdot x=\dfrac{6}{35}\)

         \(x=\dfrac{6}{35}\div\dfrac{2}{5}\)

        \(x=\dfrac{3}{7}\)

a: \(\Leftrightarrow x=\dfrac{16\cdot6}{4}=24\)

b: \(\Leftrightarrow x=\dfrac{-5\cdot42}{14}=-15\)

c: \(\Leftrightarrow\dfrac{x}{11}=\dfrac{-3}{11}\)

hay x=-3

x = 13

a) \(\dfrac{2x+5}{2x+1}=\dfrac{2x+1+4}{2x+1}=\dfrac{2x+1}{2x+1}+\dfrac{4}{2x+1}=1+\dfrac{4}{2x+1}\)  

Để \(\dfrac{2x+5}{2x+1}\in Z\) thì \(\dfrac{4}{2x+1}\in Z\) 

\(\Rightarrow4\) ⋮ \(2x+1\)

\(\Rightarrow2x+1\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)

\(\Rightarrow2x\in\left\{0;-2;1;-3;3;-5\right\}\)

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

Mà x nguyên \(\Rightarrow\text{x}\in\left\{0;-1\right\}\) 

b) \(\dfrac{3x+5}{x+1}=\dfrac{3x+3+2}{x+1}=\dfrac{3\left(x+1\right)+2}{x+1}=\dfrac{3\left(x+1\right)}{x+1}+\dfrac{2}{x+1}=3+\dfrac{2}{x+1}\) 

Để \(\dfrac{3x+5}{x+1}\in Z\) thì \(\dfrac{2}{x+1}\in Z\) 

\(\Rightarrow2\) ⋮ \(x+1\)

\(\Rightarrow x+1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)

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

c) \(\dfrac{3x+8}{x-1}=\dfrac{3x-3+11}{x-1}=\dfrac{3\left(x-1\right)+11}{x-1}=\dfrac{3\left(x-1\right)}{x-1}+\dfrac{11}{x-1}=3+\dfrac{11}{x-1}\)  

Để: \(\dfrac{3x+8}{x-1}\in Z\) thì \(\dfrac{11}{x-1}\in Z\)

\(\Rightarrow11\) ⋮ \(x-1\)

\(\Rightarrow x-1\inƯ\left(11\right)=\left\{1;-1;11;-11\right\}\)

\(\Rightarrow x\in\left\{2;0;12;-10\right\}\)

d) \(\dfrac{5x+12}{x-2}=\dfrac{5x-10+22}{x-2}=\dfrac{5\left(x-2\right)+22}{x-2}=\dfrac{5\left(x-2\right)}{x-2}+\dfrac{22}{x-2}=5+\dfrac{22}{x-2}\)

Để: \(\dfrac{5x+12}{x-2}\in Z\) thì \(\dfrac{22}{x-2}\in Z\)

\(\Rightarrow22\) ⋮ \(x-2\)

\(\Rightarrow x-2\inƯ\left(22\right)=\left\{1;-1;2;-2;11;-11;22;-22\right\}\)

\(\Rightarrow x\in\left\{3;1;4;0;13;-9;24;-20\right\}\)

e) \(\dfrac{7x-12}{x+16}=\dfrac{7x+112-124}{x+16}=\dfrac{7\left(x+16\right)-124}{x+16}=\dfrac{7\left(x+16\right)}{x+16}-\dfrac{124}{x+16}=7-\dfrac{124}{x+16}\)

Để \(\dfrac{7x-12}{x+16}\in Z\) thì \(\dfrac{124}{x+16}\in Z\) 

\(\Rightarrow124\) ⋮ \(x+16\)

\(\Rightarrow x+16\inƯ\left(124\right)=\left\{1;-1;2;-2;4;-4;31;-31;62;-62;124;-124\right\}\)

\(\Rightarrow x\in\left\{-15;-17;-14;-18;-12;-20;15;-47;46;-78;108;-140\right\}\)

x = -11

Dùng nguyên lý kẹp để tìm nghiệm nguyên em nhé .

5 < \(x^3\) - 15 < 16

5 + 15 < \(x^3\) < 16 + 15

20 < \(x^3\) < 31

⇒ 8 < \(x^3\) < 64

⇒ 2³ < \(x^3\) < 4³

Vì \(x\) là số nguyên nên \(x\) = 3

1. \(\text{6 + 2.(x - 19) = 16}.\)

\(\Leftrightarrow2.\left(x-19\right)=10.\)

\(\Leftrightarrow x-19=5.\)

\(\Leftrightarrow x=24.\)

Vậy \(x=24.\)

2. \(\text{(-240) : x – 16 = 64}.\)

\(\Leftrightarrow\left(-240\right):x=80.\)

\(\Leftrightarrow x=-3.\)

Vậy \(x=-3.\)

3. \(2x^3=16.\)

\(\Leftrightarrow x^3=8.\)

\(\Leftrightarrow x=2.\)

Vậy \(x=2.\)