Bài 1:
a, $\dfrac{x}{5}=\dfrac{-12}{15}$
$⇔x=\dfrac{-12.5}{15}=-4$
b, $\dfrac{z}{7}=\dfrac{-11}{-17}$
$⇔z=\dfrac{11.7}{17}=\dfrac{77}{17}$
c, Ta có: $\dfrac{x}{-6}=\dfrac{-2}{3}$
$⇔x=\dfrac{-2.(-6)}{3}=4$
Ta có: $\dfrac{10}{-y}=\dfrac{-2}{3}$
$⇔y=\dfrac{10.3}{2}=15$
Ta có: $\dfrac{z}{9}=\dfrac{-2}{3}$
$⇔z=\dfrac{-2.9}{3}=-6$
Bài 2:
a, $2.6=(-3).(-4)$
$⇔\dfrac{-4}{6}=\dfrac{2}{-3}$
$\dfrac{6}{-3}=\dfrac{-4}{2}$
Ngược lại nữa nhé
b, Tương tự
Bài 3:
a, Để $\dfrac{13}{x-1}∈Z$
$⇒x-1∈Ư(13)=\{-13;-1;1;13\}$
$⇔x∈\{-12;0;2;14\}$
b, Ta có: $\dfrac{x+3}{x-2}=\dfrac{x-2+5}{x-2}=1+\dfrac{5}{x-2}$
Để $\dfrac{x+3}{x-2}∈Z$
$⇒x-2∈Ư(5)=\{-5;-1;1;5\}$
$⇔x∈\{-3;1;3;7\}$