Question

[Đặng Thị Ngọc Anh]

Cho x^2+y^2+z^2=xy + yz+zx. Tính giá trị M = (x-y+1)^2019 + (y+z+1)^2020

Answer 1

[ミŇɦư Ἧσς ηgu lý ミ]

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

Answer 2

[Đặng Thị Ngọc Anh]

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