Question

[hoa ban]

x×(x +y+z)=7

y×(x+y+z)= -2

z(x+y+z)=1/2

Answer 1

[Lê Duy Khương]

Ta có 

 \(\hept{\begin{cases}x\left(x+y+z\right)=7\\y\left(x+y+z\right)=-2\\z\left(x+y+z\right)=\frac{1}{2}\end{cases}}\)

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

\(\Leftrightarrow\left(x+y+z\right)^2=\frac{11}{2}\)

\(\Leftrightarrow(x+y+z)^2=(\sqrt{\frac{11}{2}})^2\)

\(\Rightarrow\orbr{\begin{cases}x+y+z=\frac{\sqrt{22}}{2}\\x+y+z=-\frac{\sqrt{22}}{2}\end{cases}}\)

Trường hợp 1 : \(x+y+z=\frac{\sqrt{22}}{2}\)

Thay vào các biểu thức ta có

   \(x\times\frac{\sqrt{22}}{2}=7\Rightarrow x=\frac{7\sqrt{22}}{11}\)

\(y\times\frac{\sqrt{22}}{2}=-2\Rightarrow y=-\frac{2\sqrt{22}}{11}\)

\(z\times\frac{\sqrt{22}}{2}=\frac{1}{2}\Rightarrow z=\frac{\sqrt{22}}{22}\)

Trường hợp 2 : \(x+y+z=-\frac{\sqrt{22}}{2}\)

Thay vào các biểu thức ta có

\(x\times-\frac{\sqrt{22}}{2}=7\Rightarrow x=-\frac{7\sqrt{22}}{11}\)

\(y\times-\frac{\sqrt{22}}{2}=-2\Rightarrow y=\frac{2\sqrt{22}}{11}\)

\(z\times-\frac{\sqrt{22}}{2}=\frac{1}{2}\Rightarrow z=-\frac{\sqrt{22}}{22}\)

    Vậy \(x=\frac{7\sqrt{22}}{11};y=-\frac{2\sqrt{22}}{11};z=\frac{\sqrt{22}}{22}\)

           \(x=-\frac{7\sqrt{22}}{11};y=\frac{2\sqrt{22}}{11};z=-\frac{\sqrt{22}}{22}\)

Answer 2

[hoa ban]

Thanks bạn nhưng mk chưa học căn bậc 2