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\(a^{2020}+b^{2020}=a^{2021}+b^{2021}=a^{2022}+b^{2022}\) (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 :
\(a^{2020}=a^{2021}=a^{2022}\Leftrightarrow\orbr{\begin{cases}a=1\\a=0\end{cases}\Leftrightarrow}a=1\left(a>0\right)\)
\(\Rightarrow a=b=1\)\(\Rightarrow a^{2020}+b^{2021}=1^{2020}+1^{2021}=2\)
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
\(x^2+y^2+z^2=xy+yz+zx\\ \Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\\ \Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\z-x=0\end{matrix}\right.\Leftrightarrow x=y=z\\ \text{Mà }x+y+z=-3\Leftrightarrow x=y=z=-1\\ \Leftrightarrow B=1-1+1=1\)
\(2x^2+y^2+9=6x+2xy\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(x-3\right)^2=0\Leftrightarrow\hept{\begin{cases}x-3=0\\x-y=0\end{cases}}\Leftrightarrow x=y=3\)
\(\Rightarrow A=x^{2019}.y^{2020}-x^{2020}.y^{2019}+\frac{1}{9xy}=\frac{1}{27}\)
oh no bài thứ nhất là dạng chứng minh cs đúng ko ,
ko thể nào là dạng tìm a,b,c đc-.-
Ta có
\(a^{2020}+b^{2020}=a^{2021}+b^{2021}\)
\(\Leftrightarrow a^{2021}-a^{2020}=b^{2020}-b^{2021}\)
\(\Leftrightarrow a^{2020}\left(a-1\right)=b^{2020}\left(1-b\right)\)
\(\Leftrightarrow\dfrac{a-1}{1-b}=\dfrac{b^{2020}}{a^{2020}}=\left(\dfrac{b}{a}\right)^{2020}\) (1)
Ta có
\(a^{2021}+b^{2021}=a^{2022}+b^{2022}\)
\(\Leftrightarrow\dfrac{a-1}{1-b}=\left(\dfrac{b}{a}\right)^{2021}\) (2)
Từ (1) và (2) \(\Rightarrow\left(\dfrac{b}{a}\right)^{2020}=\left(\dfrac{b}{a}\right)^{2021}\)
\(\Rightarrow\dfrac{b}{a}=1\Rightarrow a=b\)
\(\Rightarrow2.a^{2020}=2.a^{2021}\Leftrightarrow a^{2020}=a^{2021}\Rightarrow a=b=1\)
\(\Rightarrow S=a^{2021}+b^{2021}=1+1=2\)