1 số chia hết cho 3,8 cs chia hết cho 24 ko. Hãy chứng minh
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\(\dfrac{x+2}{2021}+\dfrac{x+3}{2020}=\dfrac{x+2021}{2}+\dfrac{x+2020}{3}\)
\(=>\dfrac{x+2}{2021}+1+\dfrac{x+3}{2020}+1=\dfrac{x+2021}{2}+1+\dfrac{x+2020}{3}+1\)
\(=>\dfrac{x+2+2021}{2021}+\dfrac{x+3+2020}{2020}=\dfrac{x+2021+2}{2}+\dfrac{x+2020+3}{3}\)
\(=>\dfrac{x+2023}{2021}+\dfrac{x+2023}{2020}-\dfrac{x+2023}{2}-\dfrac{x+2023}{3}=0\)
\(=>(x+2023)(\dfrac{1}{2021}+\dfrac{1}{2020}-\dfrac{1}{2}-\dfrac{1}{3})=0\)
Mà \(\dfrac{1}{2021}+\dfrac{1}{2020}-\dfrac{1}{2}-\dfrac{1}{3} \ne 0\)
\(=>x+2023=0\)
\(=>x=-2023\)
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https://olm.vn/cau-hoi/trong-khong-gian-oxyz-cho-2-diem-a210-va-b040-xet-diem-s-thay-doi-luon-thuoc-truc-oz-goi-k-la-trung-diem-sb-h-la-hinh-chieu-vuong-goc-cua.6496889055212
\(y'=3x^2-2\left(m+1\right)x-\left(m^2-2m\right)\)
Hàm nghịch biến trên (1;2) khi và chỉ khi \(y'\le0;\forall x\in\left(1;2\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}3.y'\left(1\right)\le0\\3.y'\left(2\right)\le0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3-2\left(m+1\right)-\left(m^2-2m\right)\le0\\12-4\left(m+1\right)-\left(m^2-2m\right)\le0\end{matrix}\right.\)
\(\Leftrightarrow...\)
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(2x+5).11=77
= 2x + 5 = 77:11
= 2x + 5 = 7
= 2x = 7-5
= 2 x = 2
= x = 2:2
= x = 1
(2x+ 5).11 = 77
(2x+5) = 77 :11
2x + 5 = 7
2x = 7 -5
2x = 2
x = 2: 2
x = 1
3,8 hay 3 và 8?