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\(...=1+1+...+1+1\)
Số số 1 là :
\(\left(2022-2\right):2+1+1=1012\left(số\right)\)
Vậy kết quả là \(1x1012=1012\)
\(a)\dfrac{7}{8}=\dfrac{7\times9}{8\times9}=\dfrac{63}{72}\)
\(\dfrac{3}{9}=\dfrac{3\times8}{9\times8}=\dfrac{24}{72}\)
Do : \(\dfrac{63}{72}>\dfrac{24}{72}\) nên \(\dfrac{7}{8}>\dfrac{3}{9}\)
Không thì bạn có thể rút gọn 3/9 đi làm cho nó gọn ạ.
\(b)\) Ta thấy : \(\dfrac{2023}{2021}>1\) ( vì tử lớn hơn mẫu )
\(\dfrac{2021}{2022}< 1\) ( vì tử bé hơn mẫu )
Do đó : \(\dfrac{2023}{2021}>\dfrac{2021}{2022}\)
\(c)\dfrac{5}{6}=\dfrac{5\times7}{6\times7}=\dfrac{35}{42}\)
\(\dfrac{6}{7}=\dfrac{6\times6}{7\times6}=\dfrac{36}{42}\)
Do : \(\dfrac{36}{42}>\dfrac{35}{42}\) nên \(\dfrac{6}{7}>\dfrac{5}{6}\)
\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2023}\right)\\ =\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2022}{2023}\\ =\dfrac{1}{2023}\)
\(\dfrac{2021}{2022}.\dfrac{7}{16}+\dfrac{9}{16}.\dfrac{2021}{2022}=\dfrac{2021}{2022}\left(\dfrac{7}{16}+\dfrac{9}{16}\right)=\dfrac{2021}{2022}.1=\dfrac{2021}{2022}\)
\(2023\times28+2023\times34-2023\times52\)
\(=2023\times\left(28+34-52\right)\)
\(=2023\times10\)
\(=20230\)
`# \text {DNamNgV}`
`2023 \times 28 + 2023 \times 34 - 2023 \times 52`
`= 2023 \times (28 + 34 - 52)`
`= 2023 \times 10 `
`=20230`
2023×6+7×2023−2023
=2023×6+7×2023−2023x1
=2023×(6+7−1)
=2023×12
=24276
\(\dfrac{2022\times2023-1}{2023\times2021+2022}\)
= \(\dfrac{\left(2021+1\right)\times2023-1}{2023\times2021+2022}\)
= \(\dfrac{2023\times2021+2023-1}{2023\times2021+2022}\)
= \(\dfrac{2023\times2021+2022}{2023\times2021+2022}\)
= 1
2023×2021+20222022×2023−1
= (2021+1)×2023−12023×2021+20222023×2021+2022(2021+1)×2023−1
= 2023×2021+2023−12023×2021+20222023×2021+20222023×2021+2023−1
= 2023×2021+20222023×2021+20222023×2021+20222023×2021+2022
= 1