Mọi người giúp em bài này nha
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1 Are there many flowers to the right of the museum?
2 What is there next to the photocopy store?
3 My father works in a hospital in the city
4 How many people are there in Linh's family
5 My friend doesn't live in HN with his family
1 are there many flowers to the right of the museum?
2 what is there next to the photocopy store?
3 my father works in a hospital in the city
4 how many people are there in Linh's family
5 my friend doesn't live in Hanoi with his family
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\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+.....+\dfrac{1}{2021.2023}\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+....+\dfrac{2}{2021.2023}\right)\)
\(=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{2021}-\dfrac{1}{2023}\right)\)
\(=\dfrac{1}{2}.\left(1-\dfrac{1}{2023}\right)=\dfrac{1}{2}.\dfrac{2022}{2023}=\dfrac{1011}{2023}\)
Ta có A = \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{2021\cdot2023}\)
= \(\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{2021\cdot2023}\right)\)
= \(\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}+\dfrac{1}{2023}\right)\)
= \(\dfrac{1}{2}\left(1-\dfrac{1}{2023}\right)=\dfrac{1}{2}\cdot\dfrac{2022}{2023}=\dfrac{1011}{2023}\)
ây em học lớp 3 đưa bài lên lớp 3 đi sao lại đưa lên lớp 8
Lời giải:
$(x-1)^{x-2}=(x-1)^{x-4}$
$(x-1)^{x-2}-(x-1)^{x-4}=0$
$(x-1)^{x-4}[(x-1)^2-1]=0$
$\Rightarrow (x-1)^{x-4}=0$ hoặc $(x-1)^2-1=0$
Với $(x-1)^{x-4}=0\Rightarrow x=1$. Tuy nhiên khi đó số mũ $x-4=1-4=-3< 0$ nên loại
Với $(x-1)^2-1=0$
$\Rightarrow x-1=\pm 1\Rightarrow x=2$ hoặc $x=0$ (tm)