M+2019=2xy−yz−zx+2020M+2019=2xy−yz−zx+2020
=2xy−yz−zx+x2+y2+z2=2xy−yz−zx+x2+y2+z2
=(x+y−z2)2+3z24≥0=(x+y−z2)2+3z24≥0
⇒Mmin=0⇒Mmin=0 khi ⎧⎩⎨⎪⎪⎪⎪x+y−z2=03z24=0x2+y2+z2=2020{x+y−z2=03z24=0x2+y2+z2=2020
⇔⎧⎩⎨⎪⎪x+y=0z=0x2+y2=2020⇔{x+y=0z=0x2+y2=2020 ⇒⎧⎩⎨⎪⎪x=±1010−−−−√y=−xz=0
mình không hiểu ạ
x2 + y2 + z2 = xy + yz + zx
⇔ 2( x2 + y2 + z2 ) = 2( xy + yz + zx )
⇔ 2x2 + 2y2 + 2z2 = 2xy + 2yz + 2zx
⇔ 2x2 + 2y2 + 2z2 = 2xy + 2yz + 2zx
⇔ 2x2 + 2y2 + 2z2 - 2xy - 2yz - 2zx = 0
⇔ ( x2 - 2xy + y2 ) + ( y2 - 2yz + z2 ) + ( z2 - 2xz + x2 ) = 0
⇔ ( x - y )2 + ( y - z )2 + ( z - x )2 = 0
Vì : \(\hept{\begin{cases}\left(x-y\right)^2\\\left(y-z\right)^2\\\left(z-x\right)^2\end{cases}}\ge0\forall x,y,z\)=> ( x - y )2 + ( y - z )2 + ( z - x )2 ≥ 0 ∀ x, y, z
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}}\Leftrightarrow x=y=z\)
Khi đó M = ( x - y + 1 )2019 + ( y - z + 1 )2020 < đã sửa >
= ( x - x + 1 )2019 + ( y - y + 1 )2020
= 12019 + 12020
= 1 + 1 = 2
