Giải thích các bước giải:

8.Ta có:

$\dfrac{x+2}3=\dfrac{y-5}{-4}=\dfrac{z+1}5=\dfrac{2(x+2)-3(y-5)+(z+1)}{2\cdot 3-3\cdot (-4)+5} =\dfrac{2x-3y+z+20}{23}=\dfrac{72+20}{23}=4$

$\to x+2=12\to x=10$

      $y-5=-16\to y=-11$

      $z+1=20\to z=19$

9.Ta có:

$\dfrac{x+3}5=\dfrac{y-2}3=\dfrac{z-1}7=\dfrac{3(x+3)+5(y-2)-7(z-1)}{3\cdot 5+5\cdot 3-7\cdot 7}=-\dfrac{3x+5y-7z+6}{19}=-\dfrac{32+6}{19}=-2$

$\to x+3=-10, y-2=-6, z-1=-14$

$\to x=-13, y=-4, z=-13$