1. Tính nhanh
a.A= 5x[1/5+1/7] -- [2/5+2/17+6/10+9/51]
b.B=2003x14+1988+2001x2002/2002+2002x503+504x2002
c.[4,58:3,27+5,232:3,27]x4,08--4,08
d.6/11+7/17+8/25+10/17+16/11+17/25
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MH
8 tháng 12 2021
2. Tính (+5) + ( +4) = 9
3.Tính (- 9) + ( - 1) = -10
4.Tính : 28 + (-15) = 13
5.Tính (-15) + 12 = - 3
6.Tính (-25) + 25 = 0
7.Tính : 13 - 17 = - 4
8.Tính : (-78) - 9 = - 87
9.Tính : ( -24) - ( - 16) = - 8
10.Tính : 17 - (-3) = 20
11.Tính nhanh : (-21) - ( 48 + 52 - 21) = -100
8 tháng 12 2021
2, (+5)+(+4)=+9 hoạc 9 cũng được
3, (-9)+(-1)=-10
4, 28+(-15)=13
5, (-15)+12=-3
6, (-25)+25=0
7, 13-17=4
8, (-78)-9=-87
9, (-24)-(-16)=-40
10, 17-(-3)=20
11, (-21+21)-(48-52)=-4
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}`
10 tháng 1 2022
\(1.\) \(\dfrac{7}{36}-\dfrac{8}{-9}+\dfrac{-2}{3}=\dfrac{7}{36}+\dfrac{8}{9}-\dfrac{2}{3}=\dfrac{7+32-24}{36}=\dfrac{5}{12}.\)
\(2.\) \(\dfrac{-1}{2}+\dfrac{3}{7}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}=\dfrac{-9-2-7}{18}+\dfrac{4+3}{7}=\dfrac{-18}{18}+\dfrac{7}{7}=-1+1=0.\)
\(3.\) \(-\dfrac{10}{3}+\dfrac{13}{10}-\dfrac{1}{6}+\dfrac{1}{10}=\dfrac{13+1}{10}+\dfrac{-20-1}{6}=\dfrac{14}{10}+\dfrac{-21}{6}=\dfrac{7}{5}-\dfrac{7}{2}=-\dfrac{21}{10}.\)
\(4.\) \(\dfrac{10}{17}-\dfrac{5}{13}-\left(-\dfrac{7}{17}\right)-\dfrac{8}{13}+\dfrac{11}{25}=\dfrac{10+7}{17}+\dfrac{-5-8}{13}+\dfrac{11}{25}=\dfrac{17}{17}-\dfrac{13}{13}+\dfrac{11}{25}=\dfrac{11}{25}.\)
10 tháng 1 2022
1: \(=\dfrac{7}{36}+\dfrac{32}{36}+\dfrac{-24}{36}=\dfrac{15}{36}=\dfrac{5}{12}\)
2: \(=\left(-\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\dfrac{3}{7}+\dfrac{4}{7}\)
\(=\dfrac{-9-2-7}{18}+1=-1+1=0\)
3: \(=\left(-\dfrac{10}{3}-\dfrac{1}{6}\right)+\dfrac{14}{10}\)
\(=-\dfrac{21}{6}+\dfrac{14}{10}=\dfrac{-7}{2}+\dfrac{14}{10}=\dfrac{-35+14}{10}=-\dfrac{21}{10}\)
4: \(=\left(\dfrac{10}{17}+\dfrac{7}{17}\right)-\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\dfrac{11}{25}=\dfrac{11}{25}\)
TT
6
8 tháng 2 2020
Bằng 2048045
em nhờ anh của em giải hộ nên mới biết
anh của em lớp 6
CM
0
NA
1
8 tháng 5 2016
\(\frac{2003x14+1988+2001x2002}{2002+2002x503+504x2002}\)= \(\left(\frac{2003x14+1988}{2002+2002x503}\right)+\frac{2001x2002}{504x2002}\) = \(\frac{5}{168}+\frac{667}{168}\)= \(\frac{672}{168}=4\)
DD
30 tháng 3 2017
Kết quả là 98540
Nhé bạn
k mình đi
Nào tích đi
Mình cầu xin bạn đó
Goodbye
Trả lời:
a, \(A=5\times\left(\frac{1}{5}+\frac{1}{7}\right)-\left(\frac{2}{5}+\frac{2}{17}+\frac{6}{10}+\frac{9}{51}\right)\)
\(A=5\times\left(\frac{1}{5}+\frac{1}{7}\right)-\left(\frac{2}{5}+\frac{2}{17}+\frac{3}{5}+\frac{3}{17}\right)\)
\(A=5\times\left(\frac{1}{5}+\frac{1}{7}\right)-\left(\frac{5}{5}+\frac{5}{17}\right)\)
\(A=5\times\left(\frac{1}{5}+\frac{1}{7}\right)-5\times\left(\frac{1}{5}+\frac{1}{17}\right)\)
\(A=5\times\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{5}-\frac{1}{7}\right)\)
\(A=5\times\left(\frac{1}{7}-\frac{1}{17}\right)\)
\(A=5\times\frac{10}{119}\)
\(A=\frac{50}{119}\)
b, \(B=\frac{2003\times14+1988+2001\times2002}{2002+2002\times503+504\times2002}\)
\(B=\frac{\left(2002+1\right)\times14+1988+2001\times2002}{2002\times\left(1+503+504\right)}\)
\(B=\frac{2002\times14+14+1988+2001\times2002}{2002\times1008}\)
\(B=\frac{2002\times14+2002+2001\times2002}{2002\times1008}\)
\(B=\frac{2002\times\left(14+1+2001\right)}{2002\times1008}\)
\(B=\frac{2002\times2016}{2002\times1008}\)
\(B=2\)
c, Sửa dề
\(\left(4,58\div3,27+5,23\div3,27\right)\times4,08-4,08\)
\(=\left[\left(4,58+5,23\right)\div3,27\right]\times4,08-4,08\)
\(=\left(9,81\div3,27\right)\times4,08-4,08\)
\(=3\times4,08-4,08\)
\(=4,08\times\left(3-1\right)\)
\(=4,08\times2\)
\(=8,16\)
d, \(\frac{6}{11}+\frac{7}{17}+\frac{8}{25}+\frac{10}{17}+\frac{16}{11}+\frac{17}{25}\)
\(=\left(\frac{6}{11}+\frac{16}{11}\right)+\left(\frac{7}{17}+\frac{10}{17}\right)+\left(\frac{8}{25}+\frac{17}{25}\right)\)
\(=2+1+1\)
\(=4\)