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\(B=\left(x^2+y^2+4+2xy-4x-4y\right)+\left(x^2+z^2+1+2xz-2x-2z\right)+\left(y^2-4y+4\right)+4\)
\(B=\left(x+y-2\right)^2+\left(x+z-1\right)^2+\left(y-2\right)^2+4\ge4\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}x+y-2=0\\x+z-1=0\\y-2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=2\\z=1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có 1/x+1/y+1/z=0
=>1/x+1/y=-1/z
=>(1/x+1/y)^3= (-1/z)^3
=>1/x^3+1/y^3+3.1/x.1/y.(1/x+1/y) =-1/z^3
=>1/x^3+1/y^3+1/z^3= -3.1/x.1/y.(1/x+1/y) =3/(xyz) (vì 1/x+1/y=-1/z)
Mặt khác: 1/x+1/y+1/z=0
=>(xy+yz+zx)/(xyz)=0
=>xy+yz+zx=0
A=yz/x^2 +2yz + xz/y^2+ 2xz + xy/z^2+ 2 xy
=xyz/x^3+xyz/y^3+xyz/z^3 +2(xy+yz+zx) (vì x,y,z khác 0)
=xyz(1/x^3+1/y^3+1/z^3) (vì xy+yz+zx=0)
=xyz.3/(xyz) (vì 1/x^3+1/y^3+1/z^3=3/(xyz) )
=3
Vậy A=3.
\(2x^2+2y^2+z^2+2xy+2yz+2xz+32x+34y+545=0\)
\(\Leftrightarrow\left(x^2+2.x.16^2+16^2\right)+\left(y^2+2.y.17+17^2\right)+\left(x^2+y^2+z^2+2xy+2yz+2zx\right)=0\)\(\Leftrightarrow\left(x+16\right)^2+\left(y+17\right)^2+\left(x+y+z\right)^2=0\)
Ta có: \(\left\{{}\begin{matrix}\left(x+16\right)^2\ge0\forall z\\\left(y+17\right)^2\ge0\forall y\\\left(x+y+z\right)^2\ge0\forall x;y;z\end{matrix}\right.\)\(\Leftrightarrow\left(x+16\right)^2+\left(y+17\right)^2+\left(x+y+z\right)^2\ge0\forall x;y;z\)
Mà \(\Leftrightarrow\left(x+16\right)^2+\left(y+17\right)^2+\left(x+y+z\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x+16\right)^2=0\\\left(y+17\right)^2=0\\\left(x+y+z\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+16=0\\y+17=0\\x+y+z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-16\\y=-17\\x+y+z=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-16\\y=-17\\z-33=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-16\\y=17\\z=33\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=-16\\y=17\\z=33\end{matrix}\right.\)
Bạn ơi bước đầu tiên bạn viết sai rồi!!!
Phải là (x2 + 2.x.16 + 162) chứ không phải là (x2 + 2.x.162 +162)