15:15=?
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NT
1
CM
19 tháng 5 2021
\(1+\left(-5\right)+15=11\)
\(1+\left[\left(-5\right)+15\right]=11\)
\(1-\left(5-15\right)=11\)
DN
1
10 tháng 1 2022
1. \(\dfrac{-1}{7}-\dfrac{1}{8}=\dfrac{-8-7}{56}=\dfrac{-15}{56}.\)
2. \(\dfrac{15}{48}-\dfrac{5}{12}=\dfrac{5}{16}-\dfrac{5}{12}=-\dfrac{5}{48}.\)
3. \(\dfrac{-3}{15}-\left(\dfrac{-3}{25}\right)=\dfrac{-1}{5}+\dfrac{3}{25}=\dfrac{-2}{25}.\)
4. \(\dfrac{1}{15}-\dfrac{3}{5}=\dfrac{1-9}{15}=-\dfrac{8}{15}.\)
DN
1
10 tháng 1 2022
1. \(\dfrac{-1}{7}-\dfrac{1}{8}=\dfrac{-8-7}{56}=-\dfrac{15}{56}.\)
2. \(\dfrac{15}{48}-\dfrac{5}{12}=\dfrac{5}{16}-\dfrac{5}{12}=-\dfrac{5}{48}.\)
3. \(\dfrac{-3}{15}-\left(-\dfrac{3}{25}\right)=\dfrac{-1}{5}+\dfrac{3}{25}=-\dfrac{2}{25}.\)
4. \(\dfrac{1}{15}-\dfrac{3}{5}=\dfrac{1-9}{15}=-\dfrac{8}{15}.\)
DT
2
YN
21 tháng 2 2022
`Answer:`
\dfrac15+\dfrac1{5+10}+\dfrac1{5+10+15}\ +\,.\!.\!.+\ \dfrac1{5+10+15\ +\,.\!.\!.+\ 100}\\=\dfrac15+\dfrac1{5.(1+2)}+\dfrac1{5.(1+2+3)}\ +\,.\!.\!.+\ \dfrac1{5.(1+2+3\ +\,.\!.\!.+\ 20)}\\=\dfrac15\left(1+\dfrac1{1+2}+\dfrac1{1+2+3}\ +\,.\!.\!.+\ \dfrac1{1+2+3\ +\,.\!.\!.+\ 20}\right)\\=\dfrac15\bigg(\dfrac22+\dfrac26+\dfrac2{12}\ +\,.\!.\!.+\ \dfrac2{20.21}\bigg)\\=\dfrac25\left(\dfrac1{1.2}+\dfrac1{2.3}+\dfrac1{3.4}\ +\,.\!.\!.+\ \dfrac1{20.21}\right)\\=\dfrac25\left(1-\dfrac12+\dfrac12-\dfrac13+\dfrac13-\dfrac14\ +\,.\!.\!.+\ \dfrac1{20}-\dfrac1{21}\right)\\=\dfrac25\left(1-\dfrac1{21}\right)\\=\dfrac25\!\cdot\!\dfrac{20}{21}\\=\dfrac8{21}
YN
21 tháng 2 2022
`Answer:`
Mình gửi lại bài nhé. Mong lần này không bị lỗi như lần trước.
\(\dfrac15+\dfrac1{5+10}+\dfrac1{5+10+15}\ +\,.\!.\!.+\ \dfrac1{5+10+15\ +\,.\!.\!.+\ 100}\\=\dfrac15+\dfrac1{5.(1+2)}+\dfrac1{5.(1+2+3)}\ +\,.\!.\!.+\ \dfrac1{5.(1+2+3\ +\,.\!.\!.+\ 20)}\\=\dfrac15\left(1+\dfrac1{1+2}+\dfrac1{1+2+3}\ +\,.\!.\!.+\ \dfrac1{1+2+3\ +\,.\!.\!.+\ 20}\right)\\=\dfrac15\bigg(\dfrac22+\dfrac26+\dfrac2{12}\ +\,.\!.\!.+\ \dfrac2{20.21}\bigg)\\=\dfrac25\left(\dfrac1{1.2}+\dfrac1{2.3}+\dfrac1{3.4}\ +\,.\!.\!.+\ \dfrac1{20.21}\right)\\=\dfrac25\left(1-\dfrac12+\dfrac12-\dfrac13+\dfrac13-\dfrac14\ +\,.\!.\!.+\ \dfrac1{20}-\dfrac1{21}\right)\\=\dfrac25\left(1-\dfrac1{21}\right)\\=\dfrac25\!\cdot\!\dfrac{20}{21}\\=\dfrac8{21}\)
AT
1
VN
Võ Ngọc Phương
CTVHS
VIP
3 tháng 8 2023
\(\dfrac{7}{15}.\dfrac{4}{5}+\dfrac{7}{15}.\dfrac{1}{5}-\dfrac{7}{15}\)
\(=\dfrac{7}{15}.\left(\dfrac{4}{5}+\dfrac{1}{5}-1\right)\)
\(=\dfrac{7}{15}.0\)
\(=0\)
H9
HT.Phong (9A5)
CTVHS
3 tháng 8 2023
\(\dfrac{7}{15}\cdot\dfrac{4}{5}+\dfrac{7}{15}\cdot\dfrac{1}{5}-\dfrac{7}{15}\)
\(=\dfrac{7}{15}\cdot\left(\dfrac{4}{5}+\dfrac{1}{5}-1\right)\)
\(=\dfrac{7}{15}\cdot\left(1-1\right)\)
\(=\dfrac{7}{15}\cdot0\)
\(=0\)
5 + 10 = 15
5+10 =15
10+5 =15
15-5 =10
15-10 =5