X+1/2020 + X+2/2019=X-1/2023 + X-2/2024
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\(x+\left(x+1\right)+\left(x+2\right)+...+2023+2024=2024\)
\(\Rightarrow2023x+4090506=2024-2024-20232023\)
\(\Rightarrow x+4090506=-2023\)
\(\Rightarrow2023x=-2023-4090506\)
\(\Rightarrow2023x=-4092529\)
\(\Rightarrow x=-2023\).
a: \(\left|a-2b+3\right|^{2023}>=0\forall a,b\)
\(\left(b-1\right)^{2024}>=0\forall b\)
Do đó: \(\left|a-2b+3\right|^{2023}+\left(b-1\right)^{2024}>=0\forall a,b\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}a-2b+3=0\\b-1=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}b=1\\a=2b-3=2\cdot1-3=-1\end{matrix}\right.\)
Thay a=-1 và b=1 vào P, ta được:
\(P=\left(-1\right)^{2023}\cdot1^{2024}+2024=2024-1=2023\)
\(1\dfrac{1}{2}\times1\dfrac{1}{3}\times1\dfrac{1}{4}\times...\times1\dfrac{1}{2023}\times1\dfrac{1}{2024}\)
\(=\left(1+\dfrac{1}{2}\right)\times\left(1+\dfrac{1}{3}\right)\times\left(1+\dfrac{1}{4}\right)\times...\times\left(1+\dfrac{1}{2023}\right)\times\left(1+\dfrac{1}{2024}\right)\)
\(=\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times\dfrac{6}{5}\times...\times\dfrac{2024}{2023}\times\dfrac{2025}{2024}\)
\(=\dfrac{3\times4\times5\times...\times2024\times2025}{2\times3\times4\times...\times2023\times2024}\)
\(=\dfrac{2025}{2}\)
\(=1012,5\)
a) \(\left(x-2024\right)^{2023}=1\)
\(\Rightarrow\left(x-2024\right)^{2023}=1^{2023}\)
\(\Rightarrow x-2024=1\)
\(\Rightarrow x=2025\)
b) \(\left(2x-1\right)^5=32\)
\(\Rightarrow\left(2x-1\right)^5=2^5\)
\(\Rightarrow2x-1=2\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\dfrac{3}{2}\)
c) \(5< 2^x< 100\)
\(\Rightarrow4=2^2< 5< 2^x< 100< 128=2^7\)
\(\Rightarrow2< x< 7\)
Viết đề có sai không bạn:")?