Question

Cho \(x,y,z\in Q\); \(x^2+y^2+z^2=3\); \(x=\sqrt{y\left(2x-y\right)}\); \(y=\sqrt{z\left(2y-z\right)}\); \(z=\sqrt{x\left(2z-x\right)}\)

Chứng minh : \(\sqrt{\left(3-x^2\right)\left(3-y^2\right)\left(3-z^2\right)}=2\sqrt{2}.xyz\)

Answer 1

\(\Rightarrow\sqrt{y\left(2x-y\right)}.\sqrt{z\left(2y-z\right)}.\sqrt{x\left(2z-x\right)}=xyz\)

\(\Rightarrow\sqrt{xyz}.\sqrt{\left(2x-y\right)\left(2y-z\right)\left(2z-x\right)}=xyz\)

\(\Rightarrow\sqrt{\left(2x-y\right)\left(2y-z\right)\left(2z-x\right)}=\sqrt{xyz}\)

=>(2x-y)(2y-z)(2z-x)=xyz

=>(2x-y)²(2y-z)²(2z-x)²=x²y²z²

=>8(2x-y)²(2y-z)²(2z-x)²=8x²y²z²

(3-x²)(3-y²)(3-z²)

=3x²y²+3y²z²+3z²x²-x²y²z²

sau đó phân tích cái 8(2x-y)²(2y-z)²(2z-x)²

Answer 2

\(\Rightarrow\sqrt{y\left(2x-y\right)}.\sqrt{z\left(2y-z\right)}.\sqrt{x\left(2z-x\right)}=xyz\)

\(\Rightarrow\sqrt{xyz}.\sqrt{\left(2x-y\right)\left(2y-z\right)\left(2z-x\right)}=xyz\)

\(\Rightarrow\sqrt{\left(2x-y\right)\left(2y-z\right)\left(2z-x\right)}=\sqrt{xyz}\)

=>(2x-y)(2y-z)(2z-x)=xyz

=>(2x-y)²(2y-z)²(2z-x)²=x²y²z²

=>8(2x-y)²(2y-z)²(2z-x)²=8x²y²z²

(3-x²)(3-y²)(3-z²)

=3x²y²+3y²z²+3z²x²-x²y²z²

sau đó phân tích cái 8(2x-y)²(2y-z)²(2z-x)²

Answer 3

bạn viết căn bậc ở đâu ra mà hay thế