Question

[Nguyễn Thu Ngân]

x. (x+y+z)= -5 ; y. (x+y+z)= 100; z. (x+y+z)= 5

tìm x y z

Answer 1

[Nguyễn Thị Thương Hoài]

x.(x+y+z) = -5  ⇔ x + y + z = \(\dfrac{-5}{x}\) (1)

y.(x+y+z) = 100 ⇔ x + y + z = \(\dfrac{100}{y}\)  (2)

z. ( x+y+z) = 5 ⇔ x + y + z = \(\dfrac{5}{z}\) (3)

từ (1); (2); (3) ta có:

\(\dfrac{-5}{x}\) = \(\dfrac{100}{y}\) = \(\dfrac{5}{z}\) = \(\dfrac{-5+100+5}{x+y+z}\) = \(\dfrac{100}{x+y+z}\) 

⇒  x + y + z = 100 : ( \(\dfrac{-5}{x}\) ) 

⇒  x + y + z =  100. \(\dfrac{-x}{5}\) 

⇒ x + y + z = -20x (*)

thay (*) vào x.(x+y+z) = -5

ta có:  x.(-20x) = -5

                 x² . (-20) = -5

                  x² = (-5) : (-20)

                    x² = 1/4

                     x = +- 1/2

 ⇒ nếu  x= 1/2

⇒ x + y + z = -5: 1/2 = -10

 ⇒ y = 100 : (-10) = -10

⇒z =  5 : (-10)  = - 1/2 ⇒ (x,y,z)= (1/2; -10; -1/2)

x = -1/2 ⇒ x + y + z = (-5)x (-2) = 10

⇒ y =  100 : 10 = 10 

⇒ z = 5 : 10 = 1/2 ⇒ (x,y,z) =(-1/2; 10; 1/2)

vậy các cặp x,y z thỏa mãn đề bài là:

(x,y,z)= (-1/2 ; 10;  1/2);    (1/2; -10; -1/2)

Answer 2

[Trương Nhật Minh]

Ta có:

\(\left\{{}\begin{matrix}x\left(x+y+z\right)=-5\\y\left(x+y+z\right)=100\\z\left(x+y+z\right)=5\end{matrix}\right.\) \(\Rightarrow\left(x+y+z\right)\left(x+y+z\right)=-5+100+5=100\)

\(\Rightarrow x+y+z=\pm10\)

Xét \(x+y+z=10\) ta có:

\(\left\{{}\begin{matrix}x.10=-5\\y.10=100\\z.10=5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=10\\z=\dfrac{1}{2}\end{matrix}\right.\)

Xét \(x+y+z=-10\) ta có:

\(\left\{{}\begin{matrix}-10x=-5\\-10y=100\\-10z=5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-10\\z=\dfrac{-1}{2}\end{matrix}\right.\)

Đs....