\(\Leftrightarrow-12< x< 2y< 3z\)

\(\Leftrightarrow\left(x,y,z\right)\in\left\{\left(-11;-5;-3\right);\left(-10;-4;-2\right);\left(-9;-4;-2\right);\left(-8;-3;-1\right)\right\}\)

\(-\frac{1}{2}<\frac{x}{24}<\frac{y}{12}<\frac{z}{8}\Leftrightarrow-\frac{24}{48}<\frac{2x}{48}<\frac{4y}{48}<\frac{6z}{48}\)

Có thể tìm được rất nhiều các số nguyên x;y;z thỏa mãn

minh cug co y nhu bn dinh tuan viet

a: \(\dfrac{-4}{8}=\dfrac{x}{-10}=\dfrac{-7}{y}=\dfrac{z}{-24}\)

=>\(\dfrac{x}{-10}=\dfrac{-7}{y}=\dfrac{z}{-24}=\dfrac{-1}{2}\)

=>\(\left\{{}\begin{matrix}x=\left(-10\right)\cdot\dfrac{\left(-1\right)}{2}=5\\y=\dfrac{-7\cdot2}{-1}=14\\z=\dfrac{-24\cdot\left(-1\right)}{2}=\dfrac{24}{2}=12\end{matrix}\right.\)

b: \(\dfrac{-3}{6}=\dfrac{x}{-2}=\dfrac{-18}{y}=\dfrac{-z}{24}\)

=>\(\dfrac{x}{-2}=\dfrac{-18}{y}=\dfrac{z}{-24}=\dfrac{-1}{2}\)

=>\(\dfrac{x}{2}=\dfrac{18}{y}=\dfrac{z}{24}=\dfrac{1}{2}\)

=>\(x=2\cdot\dfrac{1}{2}=1;y=18\cdot\dfrac{2}{1}=36;z=\dfrac{24}{2}=12\)

Xét : \(\frac{-4}{8}=\frac{x}{-10}\Leftrightarrow8x=40\Leftrightarrow x=5\)

\(\frac{-4}{8}=\frac{-7}{y}\Leftrightarrow-4y=-56\Leftrightarrow y=14\)

\(\frac{-4}{8}=\frac{z}{24}\Leftrightarrow-96=8z\Leftrightarrow z=-12\)

\(\frac{12}{-6}=-2=\frac{-10}{5}=\frac{-6}{3}=\frac{34}{-17}=\frac{-18}{9}\)

Vậy...........

\(\frac{-24}{-6}=4=\frac{12}{3}=\frac{4}{\left(\pm1\right)^2}=\frac{\left(-2\right)^3}{-2}\)

Vậy......

Giải

x, y, z ∈ Z, ta có: −48=x−10=−7y=z−24−48=x−10=−7y=z−24

Ta có:

−48=x−10−48=x−10 nên (−4).(−10)=8.x(−4).(−10)=8.x

⇒x=−4.(−10)8=408=5⇒x=−4.(−10)8=408=5

−48=−7y−48=−7y nên (−4).y=8.(−7)(−4).y=8.(−7)

⇒y=8.(−7)−4=−56−4=14⇒y=8.(−7)−4=−56−4=14

−48=z−24−48=z−24 nên (−4).(−24)=8.z(−4).(−24)=8.z

⇒z=(−4).(−24)8=968=12

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