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\(M=x^{2023}-2023.\left(x^{2022}-x^{2021}+x^{2020}-x^{2019}+...+x^2-x\right)\)
Ta có : \(x=2022\Rightarrow x+1=2023\)
\(\Rightarrow M=x^{2023}-\left(x+1\right).\left(x^{2022}-x^{2021}+x^{2020}-x^{2019}+...+x^2-x\right)\)
\(\Rightarrow M=x^{2023}-\left(x+1\right)x^{2022}+\left(x+1\right)x^{2021}-\left(x+1\right)x^{2020}+\left(x+1\right)x^{2019}+...-\left(x+1\right)x^2+\left(x+1\right)x\)
\(\Rightarrow M=x^{2023}-x^{2023}-x^{2022}+x^{2022}+x^{2021}-x^{2021}-x^{2020}+x^{2020}+x^{2019}-x^{2019}-...-x^3-x^2+x^2+x\)
\(\Rightarrow M=x\)
\(\Rightarrow M=2022\)
Vậy \(M=2022\left(tạix=2022\right)\)
\(\left(x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-4\right)=0\)
\(\left(x-4\right)\left(x+3+x+5\right)=0\)
\(\left(x-4\right)\left(2x+8\right)=0\)
\(\left[{}\begin{matrix}x-4=0\Leftrightarrow x=4\\2x+8=0\Leftrightarrow x=-4\end{matrix}\right.\)
\(102-\left[8^2-48\right].0,5.\left(2^2.10+8\right):2^5\)
\(=102-\left[64-48\right].0,5.\left(40+8\right):32\)
\(=102-16.0,5.48:32\)
\(=102-\left(8.\dfrac{3}{2}\right)\)
\(=102-12\)
\(=90\)
\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{98.99}\)
\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{99}\)
\(=\dfrac{1}{3}-\dfrac{1}{99}\)
\(=\dfrac{33}{99}-\dfrac{1}{99}\)
\(=\dfrac{32}{99}\)
\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{98.99}\\ =\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{99}\\ =\dfrac{1}{3}-\dfrac{1}{99}\\ =\dfrac{32}{99}\)
Kq = 5/2 nha bạn câu này đã có người hỏi rồi bạn tự tham khảo nhé
