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Cho a,b,c thõa mãn : a^2 + b^2 +c^2 - ab -bc- ca = 0. Tính: P = (a-b)^2020 + (b-c)^2021 + (c-a)^2022
\(a^2+b^2+c^2-ab-bc-ac=0\)
\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\) (1)
Mà: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\)
Nên PT (1) \(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(a-c\right)^2=0\end{matrix}\right.\)
=> a = b = c
\(P=\left(a-b\right)^{2020}+\left(b-c\right)^{2021}+\left(c-a\right)^{2022}\)
\(=\left(a-a\right)^{2020}+\left(b-b\right)^{2021}+\left(c-c\right)^{2022}\)
= 0
b) \(\left(a^{2019}+b^{2019}\right)^2=\left(a^{2018}+b^{2018}\right)\left(a^{2020}+b^{2020}\right)\Leftrightarrow2a^{2019}b^{2019}=a^{2018}a^{2020}+a^{2020}b^{2018}\Leftrightarrow2ab=a^2+b^2\Leftrightarrow a=b\).
Do a, b dương nên a = b = 1.
Câu a thì bạn áp dụng BĐT Svacxo
tìm x y z thoả mãn đẳng thức 1/x2022+1/y2022+1/z2022=1/x2021+1/y2021+1/z2021=1/x2020+1/y2020+1/z2020
Đặt A = 5x2 + 2y2 + 4xy - 2x + 4y + 2022
= (2x2 + 4xy + 2y2) + 4(x + y) + 2 + (3x2 - 6x + 3) + 2017
= 2(x + y)2 + 4(x + y) + 2 + 3(x - 1)2 + 2017
= 2(x + y + 1)2 + 3(x - 1)2 + 2017 \(\ge\)2017
=> Min A = 2017
\(5x^2+2y^2+4xy-2x+4y+2022\)
\(=\left(4x^2+4x+y^2\right)+\left(y^2+4y+4\right)+\left(x^2-2x+1\right)+2017\)
\(=\left(2x+y\right)^2+\left(y+2\right)^2+\left(x-1\right)^2+2017\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}2x+y=0\\y+2=0\\x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}}\)
Vậy \(Min_A=2017\Leftrightarrow x=1;y=-2\)
\(a^{2020}+b^{2020}=a^{2021}+b^{2021}=a^{2022}+b^{2022}\) (1)
Ta có : \(a^{2021}+b^{2021}=a^{2022}+b^{2022}\)
\(\Leftrightarrow a^{2021}+b^{2021}=a^{2022}+a^{2021}b+b^{2022}+ab^{2021}-a^{2021}b-ab^{2021}\)
\(\Leftrightarrow a^{2021}+b^{2021}=a^{2021}\left(a+b\right)+b^{2021}\left(a+b\right)-ab\left(a^{2020}+b^{2020}\right)\)
\(\Leftrightarrow a^{2021}+b^{2021}=\left(a^{2021}+b^{2021}\right)\left(a+b\right)-ab\left(a^{2020}+b^{2020}\right)\)
\(\Leftrightarrow a+b-ab=1\)
\(\Leftrightarrow\left(1-b\right)\left(a-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a-1=0\\1-b=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=1\\b=1\end{cases}}}\)
(+) Thay \(a=1\)vào \(\left(1\right)\)ta được :
\(b^{2020}=b^{2021}=b^{2022}\Leftrightarrow\orbr{\begin{cases}b=0\\b=1\end{cases}\Leftrightarrow}b=1\left(b>0\right)\)
(+) Thay \(b=1\)vào (1) ta được :
\(a^{2020}=a^{2021}=a^{2022}\Leftrightarrow\orbr{\begin{cases}a=1\\a=0\end{cases}\Leftrightarrow}a=1\left(a>0\right)\)
\(\Rightarrow a=b=1\)\(\Rightarrow a^{2020}+b^{2021}=1^{2020}+1^{2021}=2\)