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1 tháng 2 2017
sau 58 ,7,50 ,2,-60 là phép tính khác .
mấy b giúp mk nha ?
HP
15 tháng 4 2018
a, \(\dfrac{5}{1.3}\)+\(\dfrac{5}{3.5}\)+\(\dfrac{5}{5.7}\)+...+\(\dfrac{5}{99.101}\)
= 5.\(\dfrac{1}{1.3}\)+5.\(\dfrac{1}{3.5}\)+5.\(\dfrac{1}{5.7}\)+...+5.\(\dfrac{1}{99.101}\)
=5.(\(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{99.101}\))
=5.(\(\dfrac{2}{2}\).\(\dfrac{1}{1.3}\)+\(\dfrac{2}{2}\).\(\dfrac{1}{3.5}\)+\(\dfrac{2}{2}\).\(\dfrac{1}{5.7}\)+...+\(\dfrac{2}{2}\).\(\dfrac{1}{99.101}\))
=\(\dfrac{5}{2}\).(\(\dfrac{2}{1.3}\)+\(\dfrac{2}{3.5}\)+\(\dfrac{2}{5.7}\)+...+\(\dfrac{2}{99.101}\))
=\(\dfrac{5}{2}\).(\(\dfrac{1}{1}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{7}\)+...+\(\dfrac{1}{99}\)-\(\dfrac{1}{101}\))
=\(\dfrac{5}{2}\).(\(\dfrac{1}{1}\)-\(\dfrac{1}{101}\))
=\(\dfrac{5}{2}\).\(\dfrac{100}{101}\)
=\(\dfrac{250}{101}\)
=\(2\dfrac{48}{101}\)
b,\(\dfrac{-11}{23}\).\(\dfrac{6}{7}\)+\(\dfrac{8}{7}\).\(\dfrac{-11}{23}\)-\(\dfrac{1}{23}\)
=\(\dfrac{-11}{23}\).(\(\dfrac{6}{7}\)+\(\dfrac{8}{7}\))-\(\dfrac{1}{23}\)
=\(\dfrac{-11}{23}\).2-\(\dfrac{1}{23}\)
=\(\dfrac{-22}{23}\)-\(\dfrac{1}{23}\)
=-1
c,\(\dfrac{2.3}{7}\)+(\(\dfrac{2}{9}\)-\(1\dfrac{1}{3}\))-\(\dfrac{5}{3}\):\(\dfrac{1}{9}\)
=\(\dfrac{6}{7}\)+(\(\dfrac{2}{9}\)-\(\dfrac{4}{3}\))-\(\dfrac{5}{3}\).9
=\(\dfrac{6}{7}\)-\(\dfrac{10}{9}\)-\(\dfrac{5}{3}\).9
=\(\dfrac{6}{7}\)-\(\dfrac{10}{9}\)-15
=\(\dfrac{54}{63}\)-\(\dfrac{70}{63}\)-\(\dfrac{945}{63}\)
=\(\dfrac{-961}{63}\)=\(-15\dfrac{16}{63}\)
d,(20+\(9\dfrac{1}{4}\)):\(2\dfrac{1}{4}\)
=(20+\(\dfrac{37}{4}\)):\(\dfrac{9}{4}\)
=20:\(\dfrac{9}{4}\)+\(\dfrac{37}{4}\):\(\dfrac{9}{4}\)
=20.\(\dfrac{4}{9}\)+\(\dfrac{37}{4}\).\(\dfrac{4}{9}\)
=\(\dfrac{80}{9}\)+\(\dfrac{37}{9}\)
=\(\dfrac{117}{9}\)
Mình làm nhiều như thế thì sẽ không tránh được lỗi sai nên mình mong bạn hãy xem lại thật kĩ nhé!
24 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
`a)`
\(\dfrac{1}{2}-\dfrac{5}{6}+\dfrac{11}{33}-\dfrac{35}{40}\)
`=`\(\dfrac{1}{2}-\dfrac{5}{6}+\dfrac{1}{3}-\dfrac{7}{8}\)
`=`\(\dfrac{12}{24}-\dfrac{20}{24}+\dfrac{8}{24}-\dfrac{21}{24}\)
`= -21/24 = -7/8`
`b)`
\(\dfrac{2}{3}\cdot1\dfrac{3}{4}-\dfrac{8}{9}-\dfrac{17}{51}-\dfrac{1}{5}\)
`=`\(\dfrac{2}{3}\cdot\dfrac{7}{4}-\dfrac{8}{9}-\dfrac{17}{51}-\dfrac{1}{5}\)
`=`\(\dfrac{7}{6}-\dfrac{8}{9}-\dfrac{17}{51}-\dfrac{1}{5}\)
`=`\(\dfrac{5}{18}-\dfrac{17}{51}-\dfrac{1}{5}\)
`=`\(-\dfrac{1}{18}-\dfrac{1}{5}=-\dfrac{23}{90}\)
`c)`
\(\dfrac{1}{2}\cdot2-2\dfrac{5}{7}+\dfrac{6}{4}-\dfrac{10}{15}\)
`=`\(1-\dfrac{19}{7}+\dfrac{6}{4}-\dfrac{10}{15}\)
`=`\(-\dfrac{12}{7}+\dfrac{6}{4}-\dfrac{10}{15}\)
`=`\(-\dfrac{3}{14}-\dfrac{10}{15}=-\dfrac{37}{42}\)
`d) `
\(\dfrac{1}{6}\cdot\dfrac{1}{11}+\dfrac{4}{11}\cdot\left(-\dfrac{1}{6}\right)+\dfrac{8}{11}\cdot\dfrac{1}{6}+\dfrac{1}{6}\cdot\dfrac{6}{11}\)
`=`\(\dfrac{1}{6}\cdot\left(\dfrac{1}{11}-\dfrac{4}{11}+\dfrac{8}{11}+\dfrac{6}{11}\right)\)
`=`\(\dfrac{1}{6}\cdot\left(\dfrac{1-4+8+6}{11}\right)\)
`=`\(\dfrac{1}{6}\cdot1=\dfrac{1}{6}\)
`e)`
\(-17\cdot\left(-23\right)+\left(-53\right)\cdot17+17\cdot14+17\cdot\left(-24\right)\)
`= 17*(23-53+14-24)`
`= 17*(-40)`
`= -680`
`f)`
\(-19\cdot218+\left(-82\right)\cdot19-533\cdot19+\left(-19\right)\cdot167\)
`= 19*(-218-82-533-167)`
`= 19*(-1000)`
`= -19000`
`g)`
\(\dfrac{2}{5}+\dfrac{3}{8}-\dfrac{11}{44}+\dfrac{9}{16}\)
`=`\(\dfrac{2}{5}+\dfrac{3}{8}-\dfrac{1}{4}+\dfrac{9}{16}\)
`=`\(\dfrac{31}{40}-\dfrac{1}{4}+\dfrac{9}{16}\)
`=`\(\dfrac{21}{40}+\dfrac{9}{16}=\dfrac{87}{80}\)
`h)`
\(\dfrac{4}{10}-1\dfrac{5}{6}\cdot2+\dfrac{7}{8}-\dfrac{1}{9}\)
`=`\(\dfrac{4}{10}-\dfrac{11}{6}\cdot2+\dfrac{7}{8}-\dfrac{1}{9}\)
`=`\(\dfrac{4}{10}-\dfrac{11}{3}+\dfrac{7}{8}-\dfrac{1}{9}\)
`=`\(-\dfrac{49}{15}+\dfrac{7}{8}-\dfrac{1}{9}\)
`=`\(-\dfrac{287}{120}-\dfrac{1}{9}=-\dfrac{901}{360}\)
`i )`
\(3\cdot\dfrac{1}{5}-\dfrac{2}{8}-\dfrac{12}{36}+\dfrac{15}{9}\)
`=`\(\dfrac{3}{5}-\dfrac{1}{4}-\dfrac{1}{3}+\dfrac{15}{9}\)
`=`\(\dfrac{7}{20}-\dfrac{1}{3}+\dfrac{15}{9}\)
`=`\(\dfrac{1}{60}+\dfrac{15}{9}=-\dfrac{33}{20}\)
`k)`
\(\dfrac{6}{8}\cdot3\dfrac{1}{2}+4\dfrac{2}{3}-\dfrac{11}{55}+\dfrac{17}{51}\)
`=`\(\dfrac{3}{4}\cdot\dfrac{7}{2}+\dfrac{14}{3}-\dfrac{1}{5}+\dfrac{17}{51}\)
`=`\(\dfrac{21}{8}+\dfrac{14}{3}-\dfrac{1}{5}+\dfrac{17}{51}\)
`=`\(\dfrac{175}{24}-\dfrac{1}{5}+\dfrac{17}{51}\)
`=`\(\dfrac{851}{120}+\dfrac{17}{51}=\dfrac{297}{40}\)
`l )`
\(\dfrac{1}{3}\cdot3\dfrac{1}{2}-4\dfrac{2}{5}-\dfrac{26}{78}+\dfrac{17}{51}\)
`=`\(\dfrac{1}{3}\cdot\dfrac{7}{2}-\dfrac{22}{5}-\dfrac{1}{3}+\dfrac{17}{51}\)
`=`\(\dfrac{1}{3}\left(\dfrac{7}{2}-1\right)-\dfrac{22}{5}+\dfrac{17}{51}\)
`=`\(\dfrac{1}{3}\cdot\dfrac{5}{2}-\dfrac{22}{5}+\dfrac{17}{51}\)
`=`\(\dfrac{5}{6}-\dfrac{22}{5}+\dfrac{17}{51}\)
`=`\(-\dfrac{107}{30}+\dfrac{17}{51}=-\dfrac{97}{30}\)
P/s: Bạn tách bài ra hỏi nhé! Và ghi đề rõ ràng chứ đừng ghi ntnay, nhiều bạn nhìn vào rất khó nhìn!
`# \text {KaizulvG}`
3 tháng 12 2018
a) 58.75+58.50-58.25
=58.(75+50-25)
=58.100
=5800
b) \(20:2^2-5^9:5^8\)
\(20:4-5\)
\(5-5=0\)
c)\(\left(5^{19}:5^{17}-4\right):7\)
\(\left(5^2-4\right):7\)
\(\left(25-4\right):7\)
\(21:7\)
=\(3\)
d) \(84:4+3^9:3^7+5^0\)
\(=21+3^2+1\)
\(=21+9+1\)
\(30+1=31\)
e) \(295-\left(31-2^2.5\right)^2\)
\(295-\left(31-4.5\right)^2\)
\(295-\left(31-20\right)^2\)
\(295-\left(11\right)^2\)
\(295-121\)
\(=174\)
f) \(11^{25}:11^{23}-3^5:\left(1^{10}\right)+2^3-60\)
\(11^2-243:1+8-60\)
\(121-243+8-60\)
\(-122+8-60\)
\(-114-60\)
\(=-174\)
HN
10 tháng 3 2018
Bài 1: Ta chỉ cần bỏ ngoặc rồi cộng hai phân số để ra kết quả là số tự nhiên là xong
Bài 2:
A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+............+\frac{1}{2003.2004}\)
A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-.............-\frac{1}{2003}+\frac{1}{2003}-\frac{1}{2004}\)
A = \(1-\frac{1}{2004}\)
A = \(\frac{2004}{2004}-\frac{1}{2004}=\frac{2003}{2004}\)
10 tháng 3 2018
Mk làm mẫu 1 bài nha
Bài 1 :
a, = (1/4+3/4) - (5/13+8/13)+2/11
= 1 - 1 + 2/11
= 2/11
Tk mk nha
18 tháng 3 2017
\(\dfrac{7}{-25}+\dfrac{18}{25}+\dfrac{4}{23}+\dfrac{5}{7}+\dfrac{19}{23}\)
=\(\left(\dfrac{-7}{25}+\dfrac{18}{25}\right)+\left(\dfrac{4}{23}+\dfrac{19}{23}\right)+\dfrac{5}{7}\)
= \(\dfrac{11}{25} +1+\dfrac{5}{7}\)
= \(1\dfrac{11}{25}+\dfrac{5}{7}\)
= \(2\dfrac{27}{175}\)
b) \(-2+\dfrac{15}{19}+\dfrac{-15}{17}+\dfrac{15}{23}+\dfrac{4}{19}\)
=\(-2+\left(\dfrac{ }{ }\right)\)
a) (32 + 23 × 5) : 7 = (9 + 8 × 5) : 7 = (9 + 40) : 7 = 49 : 7 = 7
b) (-17) + | - 5 | + (-6)+(-4) = (-17) + 5 + (-10) = (-12)+(-10) = - (12 + 10) = - 22
c) 1125 : 1123 - 35 : (110 + 23 ) + 22 + 30 × 5
= 112 - 243 : ( 1 + 8 ) + 4 + 1 × 5
= 121 - 243 : 9 + 4 + 5
= 121 - 27 + 4 + 5
= 94 + 4 + 5
= 98 + 5
= 103