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\(2a^{2020}+2b^{2020}+2c^{2020}-2\left(ab\right)^{1010}-2\left(bc\right)^{1010}-2\left(ca\right)^{1010}=0\)
\(\Leftrightarrow\left(a^{1010}-b^{1010}\right)^2+\left(b^{1010}-c^{1010}\right)^2+\left(c^{1010}-a^{1010}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^{1010}-b^{1010}=0\\b^{1010}-c^{1010}=0\\c^{1010}-a^{1010}=0\end{matrix}\right.\)
\(\Rightarrow\left|a\right|=\left|b\right|=\left|c\right|\)
Nếu đề không cho a;b;c dương thì không tính được cụ thể giá trị A
Nếu a;b;c dương thì \(a=b=c\Rightarrow A=0\)
\(a^{2020}+b^{2020}+c^{2020}=a^{1010}b^{1010}+b^{1010}c^{1010}+c^{1010}a^{1010}\)
\(\Leftrightarrow a^{2020}+b^{2020}+c^{2020}-a^{1010}b^{1010}-b^{1010}c^{1010}-c^{1010}a^{1010}=0\)
\(\Leftrightarrow2a^{2020}+2b^{2020}+2c^{2020}-2a^{1010}b^{1010}-2b^{1010}c^{1010}-2a^{1010}c^{1010}=0\)
\(\Leftrightarrow\left(a^{2020}-2a^{1010}b^{1010}+b^{2020}\right)+\left(b^{2020}-2b^{1010}c^{1010}+c^{2020}\right)+\left(c^{2020}-2a^{1010}c^{1010}+a^{2020}\right)=0\)
\(\Leftrightarrow\left(a^{1010}-b^{1010}\right)^2+\left(b^{1010}-c^{1010}\right)^2+\left(c^{1010}-a^{1010}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(a^{1010}-b^{1010}\right)=0\\b^{1010}-c^{1010}=0\\c^{1010}-a^{1010}=0\end{cases}}\Leftrightarrow a^{1010}=b^{1010}=c^{1010}\Leftrightarrow\pm a=\pm b=\pm c\)
Rồi thay :> Còn thay kiểu nào thì mình cũng hong biết :">
\(a,ĐK:x\ne2\\ b,A=\dfrac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{3}{x-2}\\ c,x=\dfrac{2021}{1010}\Leftrightarrow A=\dfrac{3}{\dfrac{2021}{1010}-\dfrac{2020}{1010}}=\dfrac{3}{\dfrac{1}{1010}}=3030\)
\(\frac{x-1009}{1010}+\frac{x-1007}{1012}=\frac{x-1010}{1009}+\frac{x-1012}{1007}\)
\(\Rightarrow(\frac{x-1009}{1010}-1)+\left(\frac{x-1007}{1012}-1\right)=\left(\frac{x-1010}{1009}-1\right)+\left(\frac{x-1012}{1007}-1\right)\)
\(\Rightarrow\frac{x-2019}{1010}+\frac{x-2019}{1012}-\frac{x-2019}{1009}-\frac{x-2019}{1007}\)
\(\Rightarrow\left(x-2019\right)\left(\frac{1}{1010}+\frac{1}{1012}-\frac{1}{1009}-\frac{1}{1007}\right)=0\)
Ta có
\(\frac{1}{1010}+\frac{1}{1012}-\frac{1}{1009}-\frac{1}{1007}\ne0\Rightarrow x-2019=0\Rightarrow x=2019\)
\(\frac{x-1009}{1010}+\frac{x-1007}{1012}=\frac{x-1010}{1009}+\frac{x-1012}{1007}\)
\(\frac{x-1009}{1010}-1+\frac{x-1007}{1012}-1=\frac{x-1010}{1009}-1+\frac{x-1012}{1007}\)\(\frac{x-2019}{1010}+\frac{x-2019}{1012}-\frac{x-2019}{1009}-\frac{x-2019}{1007}=0\)
\(\left(x-2019\right)\left(\frac{1}{1010}+\frac{1}{1012}-\frac{1}{1009}-\frac{1}{1007}\right)=0\)
1/1010 + 1/1012 - 1/1009 - 1/1007 khác 0
=> x - 2019 =0 => x = 2019
TA có \(2A=4x^2-4x-2|2x-1|+2020\)
\(=\left(2x-1\right)^2-2|2x-1|+2019\)
\(=\left(|2x-1|-1\right)^2+2018\ge2018\)
=>\(A\ge1009\)
=> MinA=2009 xảy ra khi \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
