|x-6|-3=2021
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\(\hept{\begin{cases}x-1=a\\y-2=b\\z-3=c\end{cases}}\Rightarrow a+b+c=x+y+z-6=0\).
Ta có:
\(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Leftrightarrow a+b=0\)hoặc \(b+c=0\)hoặc \(c+a=0\).
\(\Leftrightarrow\hept{\begin{cases}a=-b\\c=0\end{cases}}\)hoặc \(\hept{\begin{cases}b=-c\\a=0\end{cases}}\)hoặc \(\hept{\begin{cases}c=-a\\b=0\end{cases}}\).
Khi đó \(P=a^{2021}+b^{2021}+c^{2021}=0\).
\(M=6+6^2+6^3+...+6^{2021}\)
\(\Rightarrow6M=6^2+6^3+6^4+...+6^{2022}\)
\(\Rightarrow6M-M=6^2+6^3+6^4+...+6^{2022}-\left(6+6^2+6^3+...+6^{2021}\right)\)
\(\Rightarrow5M=6^{2022}-6\)
\(\Rightarrow M=\dfrac{6^{2022}-6}{5}\)
\(5M+6=6^x\)
\(\Rightarrow5.\dfrac{6^{2022}-6}{5}+6=6^x\)
\(\Rightarrow6^{2022}-6+6=6^x\)
\(\Rightarrow6^{2022}=6^x\)
\(\Rightarrow x=2022\)
1/ \(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
=\(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).0\)
=\(0\)
\(=\dfrac{2021}{2022}\left(\dfrac{6}{17}-\dfrac{23}{17}\right)+\dfrac{2021}{2022}=\dfrac{-2021}{2022}+\dfrac{2021}{2022}=0\)
a) \(A=3\left|2x-\dfrac{3}{2}\right|+2021^0=3\left|2x-\dfrac{3}{2}\right|+1\ge1\)
\(minA=1\Leftrightarrow2x=\dfrac{3}{2}\Leftrightarrow x=\dfrac{3}{4}\)
b) \(B=2\left|x-6\right|+3\left(2y-1\right)^2+2021^0=2\left|x-6\right|+3\left(2y-1\right)^2+1\ge1\)
\(minB=1\Leftrightarrow\) \(\left\{{}\begin{matrix}x=6\\y=\dfrac{1}{2}\end{matrix}\right.\)
\(A=3\left|2x-\dfrac{3}{2}\right|+1\ge1\\ A_{min}=1\Leftrightarrow2x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{4}\\ B=2\left|x-6\right|+3\left(2y-1\right)^2+1\ge1\\ B_{min}=1\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=\dfrac{1}{2}\end{matrix}\right.\)
\(|x-6|-3=2021\) \(|x-6|=2021+3\) \(|x-6|=2024\) \(\Rightarrow\orbr{\begin{cases}x-6=2024\\x-6=-2024\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=2024+6\\x=-2024+6\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=2030\\x=-2018\end{cases}}\)