tính
E= x10-2022x9+2022x8-2022x7+.....+2022x2-2022x+2022
biết x=2021
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\(M=\left(x^5-2021x^4\right)-\left(x^4-2021x^3\right)+\left(x^3-2021X^2\right)-\left(x^2-2021x\right)+\left(x-2021\right)-900=-900\)
Ta có: x=2021
nên x+1=2022
Ta có: \(M=x^5-2022x^4+2022x^3-2022x^2+2022x-2921\)
\(=x^5-x^4\left(x+1\right)+x^3\left(x+1\right)-x^2\left(x+1\right)+x\left(x+1\right)-2921\)
\(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-2921\)
\(=x-2921=-900\)
\(x=2021\Leftrightarrow x+1=2022\\ \Leftrightarrow P=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-x\\ P=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-x\\ P=0\)
\(P=x^5-2022x^4+2022x^3-2022x^2+2022x-2021=x^4\left(x-2021\right)-x^3\left(x-2021\right)+x^2\left(x-2021\right)-x\left(x-2021\right)+\left(x-2021\right)\)
\(=\left(x-2021\right)\left(x^4-x^3+x^2-x+1\right)\)
\(=\left(2021-2021\right)\left(x^4-x^3+x^2-x+1\right)=0\)
\(PT\Leftrightarrow2022x^2+2022x-2021x-2021=0\)
\(\Leftrightarrow2022x\left(x+1\right)-2021\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2022x-2021\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2022x-2021=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{2021}{2022}\end{matrix}\right.\)
Vậy: \(S=\left\{-1;\dfrac{2021}{2022}\right\}\)
\(Q\left(x\right)=x^{101}-2020x^{100}-2022x^{99}+2022x^{98}+x-2021\)
\(=x^{100}\left(x-2021\right)+x^{99}\left(x-2021\right)-x^{98}\left(x-2021\right)+x^{98}+x-2021\)
\(Q\left(2021\right)=0+0-0+2021^{98}+0=2021^{98}\)
a: \(2022^{x-2021}+3=\left(7-5\right)^2\)
=>\(2022^{x-2021}+3=4\)
=>\(2022^{x-2021}=1\)
=>x-2021=0
=>x=2021
b: \(\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=795\)
=>\(30x+\left(1+2+3+...+30\right)=795\)
=>\(30x+\dfrac{30\cdot31}{2}=795\)
=>\(30x=795-31\cdot15=330\)
=>x=11
Khi x = 2021
=> 2022 = x + 1
Khi đó E = x10 - 2022x9 + 2022x8 - ... + 2022x2 - 2022x + 2022
= x10 - (x + 1)x9 + (x + 1)x8 - .... + (x + 1)x2 - (x + 1)x + (x + 1)
= x10 - x10 - x9 + x9 + x8 - ... + x3 + x2 - x2 - x + x + 1
= 1