K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
31 tháng 3 2023
1: =72/90+65/90=137/90
2: =24/56-77/56=-53/56
3: =-7/10+4/5=1/10
4: =15/100-4/100=11/100
5: =4/6-5/6=-1/6
6: =10/40-15/40-76/40=-81/40
31 tháng 3 2023
7: =-9/10+7/18
=-81/90+35/90=-46/90=-23/45
8: =27/90-55/90=-28/90=-14/45
9: =36/60-50/60-35/60=-49/60
10: =-4/9+5/6-3/8
=-32/72+60/72-27/72
=1/72
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
10 tháng 7
\(\dfrac{4}{5}\) : (\(\dfrac{4}{5}\) .- \(\dfrac{5}{4}\)) : (\(\dfrac{16}{25}\) - \(\dfrac{1}{5}\))
= \(\dfrac{4}{5}\) : (-1) : (\(\dfrac{16}{25}\) - \(\dfrac{5}{25}\))
= -\(\dfrac{4}{5}\) : \(\dfrac{11}{25}\)
= - \(\dfrac{4}{5}\) x \(\dfrac{25}{11}\)
= - \(\dfrac{20}{11}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
10 tháng 7
\(\dfrac{4}{5}\): (\(\dfrac{4}{5}\).-\(\dfrac{5}{4}\)) : (\(\dfrac{16}{25}\) - \(\dfrac{1}{5}\))
=\(\dfrac{4}{5}\) x - 1: (\(\dfrac{16}{25}\) - \(\dfrac{5}{25}\))
= - \(\dfrac{4}{5}\) : \(\dfrac{11}{25}\)
= - \(\dfrac{4}{5}\) x \(\dfrac{25}{11}\)
= - \(\dfrac{20}{11}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
10 tháng 7
\(\dfrac{11}{12}\): (\(\dfrac{7}{9}\) + - \(\dfrac{1}{3}\)) - (\(\dfrac{2}{3}\) - \(\dfrac{5}{15}\))
= \(\dfrac{11}{12}\) : (\(\dfrac{7}{9}\) - \(\dfrac{3}{9}\)) - (\(\dfrac{2}{3}\) - \(\dfrac{1}{3}\))
= \(\dfrac{11}{12}\) : \(\dfrac{4}{9}\) - \(\dfrac{1}{3}\)
= \(\dfrac{11}{12}\) x \(\dfrac{9}{4}\) - \(\dfrac{1}{3}\)
= \(\dfrac{99}{48}\) - \(\dfrac{16}{48}\)
= \(\dfrac{83}{48}\)
LM
Lê Minh Vũ
CTVHS
VIP
14 tháng 6 2023
\(\left(-\dfrac{1}{2}\right)^2\div\dfrac{1}{4}-2\times\left(-\dfrac{1}{2}\right)^2\\= \dfrac{1}{4}\div\dfrac{1}{4}-2\times\dfrac{1}{4}\\ =1-\dfrac{1}{2}\\ =\dfrac{1}{2}\)
\(\left(-2\right)^3\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right)\div\dfrac{5}{12}\)
= \(-6\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-\dfrac{11}{6}\right)\div\dfrac{5}{12}\)
= \(\dfrac{1}{4}+-\dfrac{1}{2}\div\dfrac{5}{12}\)
= \(\dfrac{1}{4}+-\dfrac{6}{5}\)
= \(\dfrac{1}{4}-\dfrac{6}{5}\)
= \(-\dfrac{19}{20}\)
\(\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\\ =\dfrac{58}{9}+\dfrac{7}{11}-\dfrac{40}{9}+\dfrac{26}{11}\\ =\dfrac{58}{9}-\dfrac{40}{9}+\dfrac{7}{11}+\dfrac{26}{11}\\ =12+3\\ =15\)
TA
14 tháng 6 2023
\(a,\left(\dfrac{-1}{2}\right)^2:\dfrac{1}{4}-2\left(-\dfrac{1}{2}\right)^2\)
\(=\left(-\dfrac{1}{2}\right)^2\left(4-2\right)\)
\(=\dfrac{1}{4}.2=\dfrac{1}{2}\)
\(b,\left(-2\right)^3.\dfrac{-1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right):\dfrac{5}{12}\)
\(=\left(-8\right).\dfrac{-1}{24}+\left(-\dfrac{1}{2}\right).\dfrac{12}{5}\)
\(=\dfrac{1}{3}+\left(-\dfrac{1}{5}\right)=\dfrac{2}{15}\)
\(c,\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\)
\(=\dfrac{701}{99}-\dfrac{206}{99}=\dfrac{495}{99}=5\)
\(d,10\dfrac{1}{5}-5\dfrac{1}{2}.\dfrac{60}{11}+\dfrac{3}{15\%}\)
\(=\dfrac{51}{5}-30+20=\dfrac{1}{5}\)
\(e,\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)
\(=\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}.\left(-\dfrac{7}{11}\right)\)
\(=-\dfrac{5}{11}\)
\(f,\dfrac{-5}{7}.\dfrac{2}{11}+\left(-\dfrac{5}{7}\right).\dfrac{9}{11}+1\dfrac{5}{7}\)
\(=\left(-\dfrac{5}{7}\right)\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\)
\(=\left(-\dfrac{5}{7}\right)+\dfrac{12}{7}=1\)
23 tháng 11 2016
1+(-2)+3+(-4)+5+(-6)+7+(-8)+9+(-10)+11+(-12)
=(1+3+5+7+9+11)+[(-2)+(-4)+(-6)+(-8)+(-10)+(-12)]
= 36+-42
=-6
23 tháng 11 2016
(-1)+2+(-3)+4+(-5)+6+(-7)+8+(-9)+10+(-11)+12
=[(-1)+(-3)+(-5)+(-7)+(-9)+(-11)]+(2+4+6+8+10+12)
=(-36)+42
=6
VT
1
NB
20 tháng 3 2023
\(-\dfrac{5}{6}-\dfrac{3}{10}=-\dfrac{17}{15}\)
\(-\dfrac{1}{3}+-\dfrac{3}{4}-\dfrac{2}{5}=-\dfrac{13}{12}-\dfrac{2}{5}=-\dfrac{89}{60}\)
\(-\dfrac{2}{9}-\dfrac{11}{6}-\dfrac{7}{12}=-\dfrac{37}{18}-\dfrac{7}{12}=-\dfrac{95}{36}\)
\(\dfrac{2}{5}--\dfrac{7}{10}-\dfrac{4}{15}=\dfrac{11}{10}-\dfrac{4}{15}=\dfrac{5}{6}\)
\(#YH\)
\(#NBaoNgoc\)
Q
2 tháng 4 2017
Bài 1: Tính
\(\text{1)}\) \(\dfrac{5}{8}.\dfrac{7}{30}-\dfrac{5}{2}.\dfrac{1}{8}\)
\(=\dfrac{5}{8}.\dfrac{7}{30}-\dfrac{5}{8}.\dfrac{1}{2}\)
\(=\dfrac{5}{8}.\left(\dfrac{7}{30}-\dfrac{1}{2}\right)\)
\(=\dfrac{5}{8}.\dfrac{-4}{15}\)
\(=\dfrac{-1}{6}\)
\(\text{2)}\) \(\dfrac{21}{10}.\dfrac{3}{4}-\dfrac{21}{10}-\dfrac{3}{4}\)
\(=\dfrac{63}{40}-\dfrac{21}{10}-\dfrac{3}{4}\)
\(=\dfrac{-21}{40}-\dfrac{3}{4}\)
\(=\dfrac{-51}{40}\)
\(\text{3)}\) \(\dfrac{-4}{11}:\dfrac{-6}{11}\)
\(=\dfrac{-4}{11}.\dfrac{11}{-6}\)
\(=\dfrac{4}{6}\)
\(\text{4)}\) \(\dfrac{2}{7}.\dfrac{14}{3}-1\)
\(=\dfrac{4}{3}-1\)
\(=\dfrac{1}{3}\)
\(\text{5)}\) \(\dfrac{4}{7}:\left(\dfrac{1}{5}.\dfrac{4}{7}\right)\)
\(=\dfrac{4}{7}:\dfrac{1}{5}:\dfrac{4}{7}\)
\(=1:\dfrac{1}{5}\)
\(=5\)
\(\text{6)}\) \(\dfrac{12}{7}.\dfrac{7}{4}+\dfrac{35}{11}:\dfrac{245}{121}\)
\(=3+\dfrac{35}{11}.\dfrac{121}{245}\)
\(=3+\dfrac{11}{7}\)
\(=3\dfrac{11}{7}=\dfrac{32}{7}\)
\(\text{7)}\) \(\left(\dfrac{4}{3}+\dfrac{8}{3}\right).\left(\dfrac{7}{4}-\dfrac{6}{4}\right):\left(\dfrac{6}{5}+\dfrac{12}{5}+\dfrac{1}{5}\right)\)
\(=4.\left(\dfrac{7}{4}-\dfrac{6}{4}\right):\left(\dfrac{6}{5}+\dfrac{12}{5}+\dfrac{1}{5}\right)\)
\(=4.\dfrac{1}{4}:\left(\dfrac{6}{5}+\dfrac{12}{5}+\dfrac{1}{5}\right)\)
\(=4.\dfrac{1}{4}:\dfrac{19}{5}\)
\(=1:\dfrac{19}{5}\)
\(=\dfrac{5}{19}\)
\(\text{8)}\) \(\left(\dfrac{1}{4}-\dfrac{1}{4}+\dfrac{\dfrac{1}{9}}{\dfrac{1}{9}}\right):\left(\dfrac{2}{3}+\dfrac{\dfrac{7}{15}}{\dfrac{2}{5}}-\dfrac{1}{6}\right)\)
\(=\left(0+1\right):\left(\dfrac{2}{3}+\dfrac{7}{15}:\dfrac{2}{5}-\dfrac{1}{6}\right)\)
\(=1:\left(\dfrac{2}{3}+\dfrac{7}{6}-\dfrac{1}{6}\right)\)
\(=1:\left(\dfrac{2}{3}+1\right)\)
\(=1:\dfrac{5}{3}\)
\(=\dfrac{3}{5}\)
\(\text{9)}\)
\(\left[\left(\dfrac{2}{193}-\dfrac{3}{389}\right).\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}-\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\left[\dfrac{199}{75077}.\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}-\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\left[\dfrac{199}{6613}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}-\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\dfrac{13235}{13226}:\left[\left(\dfrac{7}{1931}-\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\dfrac{13235}{13226}:\left[\dfrac{3}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\dfrac{13235}{13226}:\left[\dfrac{3}{50}+\dfrac{9}{2}\right]\)
\(=\dfrac{13235}{13226}:\dfrac{114}{25}\)
\(=\dfrac{330875}{1507764}\)
\(S=\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+\frac{1}{4\cdot5\cdot6}+...+\frac{1}{10\cdot11\cdot12}\)
\(2S=\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+\frac{2}{4\cdot5\cdot6}+...+\frac{2}{10\cdot11\cdot12}\)
\(2S=\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+\left(\frac{1}{3\cdot4}-\frac{1}{4\cdot5}\right)+\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}\right)+...+\left(\frac{1}{10\cdot11}-\frac{1}{11\cdot12}\right)\)
\(2S=\frac{1}{2\cdot3}-\frac{1}{11\cdot12}\)
\(2S=\frac{1}{6}-\frac{1}{132}\)
\(2S=\frac{21}{132}\)
\(S=\frac{21}{132}\div2\)
\(S=\frac{21}{264}\)