ta có

x.y.y.z.x.z =1/3.(-2/5).(-3/10)=1/25

nên (x.y.z)^2 =1/25

+) x.y.z=1/5 nên x= 1/5:1/3=3/5

                        y=1/5:(-2/5)=-1/2

                        z=1/5:(-3/10)=-2/3

+)x.y.z = -1/5 nên x=-1/5 :1/3 =-3/5

y= -1/5:(-2/5) =1/2

z=-1/5:(-3/10)=2/3.

sau đó bạn tự kết luận nhé

Từ đề bài ta có: \(\left(x.y.z\right)^2=\frac{1}{3}.\frac{-2}{5}.\frac{-3}{10}=\frac{1}{25}\Rightarrow\orbr{\begin{cases}xyz=\frac{1}{5}\\xyz=-\frac{1}{5}\end{cases}}\)

Với \(xyz=\frac{1}{5}\Rightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=-\frac{2}{3}\\z=\frac{3}{5}\end{cases}}\)

Với \(xyz=\frac{-1}{5}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{3}\\z=\frac{-3}{5}\end{cases}}\)

\(x^2=y.z\Rightarrow x^3=x.y.z\\ y^2=x.z\Rightarrow y^3=x.y.z\\ z^2=x.y\Rightarrow z^3=x.y.z\\ \Rightarrow x^3=y^3=z^3\\ \Rightarrow x=y=z\)

Ta có:

x.y = z (1)

y.z = 4.x (2)

x.z = 4.y (3)

Từ (1), (2) và (3) => (x.y).(y.z).(x.z) = z.(4.x).(4.y)

=> (x.y.z)² = 16.x.y.z

=> (x.y.z)² - 16.x.y.z = 0

=> x.y.z.(x.y.z - 16) = 0

\(\Rightarrow\left[\begin{array}{nghiempt}x.y.z=0\\x.y.z-16=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x.y.z=0\\x.y.z=16\end{array}\right.\)

+ Với x.y.z = 0 => \(\left[\begin{array}{nghiempt}x=0\\y=0\\z=0\end{array}\right.\)

+ Với x.y.z = 16 => x.y = \(\frac{16}{z}\) = z (từ (1)) => z² = 16 => \(z\in\left\{4;-4\right\}\)

Tương tự với (2) và (3) ta được 4 cặp giá trị (x;y;z) tương ứng thỏa mãn là: (2;2;4) ; (-2;-2;4) ; (-2;2;-4) ; (2;-2;-4)

Vậy ...

a,Ta có: x+y= -7/6 và y+z= 1/4

=>x+y+y+z= -7/6 +1/4

=>x+z+2y= -11/12

=>1/2+2y= -11/12

=>2y= -11/12 -1/2

=>2y= -17/12

=>y= -17/24

Mà x+y=-7/6 =>x= -7/6+17/24= -11/24

      x+z=1/2 =>z=1/2+11/24=23/24

Ta có: \(x+y=-\frac{7}{6};y+z=\frac{1}{4};x+z=\frac{1}{2}\)

\(\Rightarrow\left(x+y\right)+\left(y+z\right)+\left(x+z\right)=-\frac{7}{6}+\frac{1}{4}+\frac{1}{2}\)

\(\Rightarrow2x+2y+2z=-\frac{28}{24}+\frac{6}{24}+\frac{12}{24}\)

\(\Rightarrow2\left(x+y+z\right)=-\frac{5}{12}\)

\(\Rightarrow x+y+z=-\frac{5}{12}:2\)

\(\Rightarrow x+y+z=-\frac{5}{24}\)

\(\Rightarrow\left(x+y+z\right)-\left(x+y\right)=-\frac{5}{24}+\frac{7}{6}\Rightarrow z=-\frac{5}{24}+\frac{28}{24}=\frac{23}{24}\)

\(\Rightarrow\left(x+y+z\right)-\left(y+z\right)=-\frac{5}{24}-\frac{1}{4}\Rightarrow x=-\frac{5}{24}-\frac{6}{24}=-\frac{11}{24}\)

\(\Rightarrow\left(x+y+z\right)-\left(x+z\right)=-\frac{5}{24}-\frac{1}{2}\Rightarrow y=-\frac{5}{24}-\frac{12}{24}=-\frac{17}{24}\)

Vậy \(x=\frac{23}{24};y=-\frac{17}{24};z=-\frac{11}{24}\)

Chuk pạn hok tốt!

b,Ta có:   x.y=1/3 và y.z= -2/5

=>x.y.y.x=1/3.(-2/5)

=>x.z.y^2= -2/15

=>-3/10.y^2= -2/15

=>y^2=4/9

=>y=2/3

  Mà x.y=1/3 =>x=1/3:2/3=1/2

       x.z= -3/10 =>z= -3/10:1/2 = -3/5

\(x^2=y.z\Rightarrow\frac{x}{y}=\frac{z}{x}\)

tuong tự ta có\(\frac{x}{y}=\frac{z}{x}=\frac{y}{z}=\frac{x+z+y}{y+x+z}=1\)

=> dpcm

Lile nhá bạn