6.
a) x:y=2:3 ⇔ x=2/3 .y
Lại có: x-y=-2⇔ 2y/3-y=-2⇔-y/3=-2⇔y=6
⇒x=2/3 .6 =4
Vậy x=4; y=6
b)x=3/4 .y ⇒xy=3/4 . $y^2$=48
⇔$y^2$=64 ⇔\(\left[ \begin{array}{l}y=8⇒x=6\\y=-8⇒x=-6\end{array} \right.\)
7.
a) x=2/5 .y ; z= 4/3 . y
⇒x+y+z= 2/5 .y +y+ 4/3 . y=-82
⇔41/15.y=82
⇔y=30
⇒x=12;z=40
b) Ta có:
$(x-1)^2≥0$
$(y-1/2)^2≥0$
$(z-2020)^2≥0$
⇒$(x-1)^2+(y-1/2)^2+(z-2020)^2≥0$
Vậy pt⇔ $(x-1)^2=0$;$(y-1/2)^2=0$;$(z-2020)^2=0$
⇔x=1; y=1/2; z=2020
