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\(\Rightarrow\sqrt{y\left(2x-y\right)}.\sqrt{z\left(2y-z\right)}.\sqrt{x\left(2z-x\right)}=xyz\)
\(\Rightarrow\sqrt{xyz}.\sqrt{\left(2x-y\right)\left(2y-z\right)\left(2z-x\right)}=xyz\)
\(\Rightarrow\sqrt{\left(2x-y\right)\left(2y-z\right)\left(2z-x\right)}=\sqrt{xyz}\)
=>(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
\(\Rightarrow\sqrt{y\left(2x-y\right)}.\sqrt{z\left(2y-z\right)}.\sqrt{x\left(2z-x\right)}=xyz\)
\(\Rightarrow\sqrt{xyz}.\sqrt{\left(2x-y\right)\left(2y-z\right)\left(2z-x\right)}=xyz\)
\(\Rightarrow\sqrt{\left(2x-y\right)\left(2y-z\right)\left(2z-x\right)}=\sqrt{xyz}\)
=>(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
Vì \(\sqrt{\left(x-\sqrt{2}\right)^2}=\left|x-\sqrt{2}\right|\ge0;\sqrt{\left(y+\sqrt{2}\right)^2}=\left|y+\sqrt{2}\right|\ge0\);|x+y+z|\(\ge\)0
=>\(\left|x-\sqrt{2}\right|+\left|y+\sqrt{2}\right|+\left|x+y+z\right|\ge0\)
Dấu "=" xảy ra khi \(\left|x-\sqrt{2}\right|=\left|y+\sqrt{2}\right|=\left|x+y+z\right|=0\)
\(\left|x-\sqrt{2}\right|=0\Leftrightarrow x-\sqrt{2}=0\Leftrightarrow x=\sqrt{2}\)
\(\left|y+\sqrt{2}\right|=0\Leftrightarrow y+\sqrt{2}=0\Leftrightarrow y=-\sqrt{2}\)
\(\left|x+y+z\right|=0\Leftrightarrow x+y+z=0\Leftrightarrow\sqrt{2}+\left(-\sqrt{2}\right)+z=0\Leftrightarrow z=0\)
Vậy ............
3: |2x-1|=|x+1|
=>2x-1=x+1 hoặc 2x-1=-x-1
=>x=2 hoặc 3x=0
=>x=2 hoặc x=0
4: \(\Leftrightarrow\left\{{}\begin{matrix}x+\sqrt{5}=0\\y-\sqrt{3}=0\\x-y-z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\sqrt{5}\\y=\sqrt{3}\\z=x-y=-\sqrt{5}-\sqrt{3}\end{matrix}\right.\)
\(\sqrt{\left(2x-\sqrt{16}\right)^2}+\left(y^2.64\right)^2+lx+y+zl=0\)
\(\Rightarrow\sqrt{2x-4}+8y^4+lx+y+zl=0\)
\(\sqrt{2x-4};8y^4;lx+y+zl\ge0\)mà \(\sqrt{2x-4}+8y^4+lx+y+zl=0\)
\(\Rightarrow\sqrt{2x-4}=8y^4=lx+y+zl=0\)
=>2x-4=y4=lx+y+zl=0
=>x=2;y=0;z=-2
Vậy x=2;y=0;z=-2
vô yahoo hỏi đáp là biết