Vị trí tương đối của hai mặt cầu (S) có tâm I(1;1;1), bán kính R = 1 và mặt cầu (S’) có tâm I'(3;3;3), bán kính R’=1 là:

A. ${\mathrm{x}}^2 + {\mathrm{y}}^2 + {\mathrm{z}}^2$ - 2x + 4y - 8z  + 25 = 0

B. ${\mathrm{x}}^2 + {\mathrm{y}}^2 + {\mathrm{z}}^2$ - 2x - 4y - 6z + 15 = 0

C. 3${\mathrm{x}}^2 +3 {\mathrm{y}}^2 + 3{\mathrm{z}}^2$ - 6x - 7y - 8z + 1 = 0

D. ${(\mathrm{x} - 1)}^2 + {(\mathrm{y} + 2)}^2 + {(\mathrm{z} + 3)}^2$ + 10 = 0

A. ở ngoài nhau

B. tiếp xúc

C. cắt nhau

D. chứa nhau

A. A(1; 2; 3), B(5; -4; -1), C(3; -1; 1)

B. A(1; 2; 3), B(5; -4; -1), C(6; -2; 2)

C. A(1; 2; 3), B(5; -4; -1), C(9; -10; -5)

D. A(1; 2; 3), B(5; -4; -1), C(-3; 8; 7)

A. $\mathrm{I}\left(-1;\frac{4}{3};-\frac{5}{2}\right), \mathrm{R}=\frac{361}{36}$

B. $\mathrm{I}\left(-1;\frac{4}{3};-\frac{5}{2}\right), \mathrm{R}=\frac{19}{6}$

C. $\mathrm{I}\left(-3;4;-\frac{15}{2}\right), \mathrm{R}=\frac{19}{6}$

D. $\mathrm{I}\left(3;-4;\frac{15}{2}\right), \mathrm{R}=\frac{361}{36}$

A. ${(\mathrm{x} - 1)}^2 + {(\mathrm{y} + 2)}^2 + {(\mathrm{z} + 3)}^2$ = 3

B. ${(\mathrm{x} + 1)}^2 + {(\mathrm{y} - 2)}^2 + {(\mathrm{z} - 3)}^2$ = 9

C. ${(\mathrm{x} + 1)}^2 + {(\mathrm{y} - 2)}^2 + {(\mathrm{z} - 3)}^2$ = 3

D. ${(\mathrm{x} - 1)}^2 + {(\mathrm{y} + 2)}^2 + {(\mathrm{z} + 3)}^2$ = 9

A. ${\mathrm{x}}^2 + {\mathrm{y}}^2 + {\mathrm{z}}^2$ + 2x - y - 2z = 0

B. ${\mathrm{x}}^2 + {\mathrm{y}}^2 + {\mathrm{z}}^2$ + 4x + 2y - 4z = 0

C. ${\mathrm{x}}^2 + {\mathrm{y}}^2 + {\mathrm{z}}^2$ + 4x - 2y + 4z = 0

D. ${\mathrm{x}}^2 + {\mathrm{y}}^2 + {\mathrm{z}}^2$ + 4x - 2y - 4z = 0

A. $\mathrm{I}\left(1;-1;\frac{3}{2}\right), \mathrm{R}=9$

B. $\mathrm{I}\left(2;-2;3\right), \mathrm{R}=\frac{9}{2}$

C. $\mathrm{I}\left(1;-1;\frac{3}{2}\right), \mathrm{R}=\frac{9}{2}$

D. $\mathrm{I}\left(2;-2;3\right), \mathrm{R}=9$