Ta có: 

  x² + 2y² + z² − 2xy − 2yz + xz − 3x − z + 5 = 0

<=>\(\left(x-\frac{2y+3}{2}\right)^2\) + \(\left(y-\frac{z+3}{2}\right)^2\)+ \(\frac{1}{2}\).( z - 1 )²=0

<=> \(\hept{\begin{cases}x=3\\y=2\\z=1\end{cases}}\)

Do đó: S= 3³ + 2^{7 +} 1²⁰¹⁰ = 156

-C.ơn bạn nha ^^ :))

b, x² +y²+z² +2x-4y-6z+14=0

<=> (x²+2x+1)+(y²-4y+4)+(z²-6z+9)=0

<=> (x+1)²+(y-2)²+(z-3)²=0

=>(x+1)²=(y-2)²=(z-3)²=0

=>x+1=y-2=z-3=0

=> x=-1; y=2; z=3

c, 2x²+y²-6x-4y+2xy+5=0

<=> (x²+y²+4+2xy-4x-4y)+(x²-2x+1)=0

<=> (x+y-2)²+(x-1)²=0

=> (x+y-2)²=(x-1)²=0

=>x+y-2=x-1=0

=>x=1; y=1

Bài này ez thôi, làm mãi rồi.

Theo đề bài, ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)

=>\(\dfrac{xy+yz+xz}{xyz}=0\)

=> xy+yz+zx=0

=> \(\left\{{}\begin{matrix}xy=-yz-zx\\yz=-xy-zx\\zx=-xy-yz\end{matrix}\right.\)

Ta có: x²+2yz=x²+yz-xy-zx=(x-y)(x-z)

           y²+2xz=y²+xz-xy-yz=(x-y)(z-y)

           z²+2xy=z²+xy-yz-xz=(x-z)(y-z)

=> \(\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(x-y\right)\left(z-y\right)}+\dfrac{xy}{\left(x-z\right)\left(y-z\right)}=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\dfrac{\left(x-y\right)\left(x-z\right)\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)

bí à bạn