1 bỏ ngoặc rồi tính
\(\dfrac{3}{2}+\left(\dfrac{1}{5}-\dfrac{3}{4}\right)\)
\(1\dfrac{1}{3}+\left(\dfrac{2}{3}-\dfrac{3}{4}\right)-\left(0,8+1\dfrac{1}{5}\right)\)
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1: Ta có: \(23\dfrac{1}{4}\cdot\dfrac{7}{5}-13\dfrac{1}{4}:\dfrac{5}{7}\)
\(=\dfrac{93}{4}\cdot\dfrac{7}{5}-\dfrac{53}{4}\cdot\dfrac{7}{5}\)
\(=\dfrac{7}{5}\cdot10=14\)
2: Ta có: \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)
\(=\dfrac{12+8-3}{12}\cdot\dfrac{1}{400}\)
\(=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\)
a: \(x=\left(-\dfrac{2}{3}\right)^5:\left(-\dfrac{2}{3}\right)^2=\left(-\dfrac{2}{3}\right)^3=-\dfrac{8}{27}\)
b: =>x-1/2=1/3
=>x=5/6
c: =>2/3x-1=0 hoặc 3/4x+1/2=0
=>x=3/2 hoặc x=-1/2:3/4=-1/2*4/3=-4/6=-2/3
d =>4/9:x=10/3:9/4=10/3*4/9=40/27
=>x=4/9:40/27=4/9*27/40=108/360=3/10
1: \(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\)
\(=\left(\dfrac{11}{24}+\dfrac{13}{24}\right)-\left(\dfrac{5}{41}+\dfrac{36}{41}\right)+\dfrac{1}{2}\)
\(=1-1+\dfrac{1}{2}=\dfrac{1}{2}\)
2: \(12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)
\(=12:\left(\dfrac{9}{12}-\dfrac{10}{12}\right)^2\)
\(=12:\left(-\dfrac{1}{12}\right)^2=12:\dfrac{1}{144}=12\cdot144=1368\)
3: \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\cdot\left(0,8-\dfrac{3}{4}\right)^2\)
\(=\dfrac{12+8-3}{12}\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)
\(=\dfrac{17}{12}\cdot\left(\dfrac{16-15}{20}\right)^2\)
\(=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\)
4: \(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\dfrac{3}{5}\)
\(=\dfrac{5}{3}\cdot\left(-16-\dfrac{2}{7}\right)+\dfrac{5}{3}\cdot\left(28+\dfrac{2}{7}\right)\)
\(=\dfrac{5}{3}\left(-16-\dfrac{2}{7}+28+\dfrac{2}{7}\right)\)
\(=12\cdot\dfrac{5}{3}=20\)
5: \(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)
\(=\left(4\cdot\dfrac{3}{4}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)
\(=\dfrac{5}{2}\cdot\dfrac{6}{5}-17=3-17=-14\)
6: \(\left(\dfrac{1}{3}\right)^{50}\cdot\left(-9\right)^{25}-\dfrac{2}{3}:4\)
\(=\left(\dfrac{1}{3}\right)^{50}\cdot\left(-1\right)\cdot3^{50}-\dfrac{2}{3\cdot4}\)
\(=-1-\dfrac{2}{12}=-1-\dfrac{1}{6}=-\dfrac{7}{6}\)
`1//([-1]/2)^2 . |+8|-(-1/2)^3:|-1/16|=1/4 .8+1/8 .16=2+2=4`
`2//|-0,25|-(-3/2)^2:1/4+3/4 .2017^0=0,25-2,25.4+0,75.1=0,25-9+0,75=-8,75+0,75-8`
`3//|2/3-5/6|.(3,6:2 2/5)^3=|-1/6|.(3/2)^3=1/6 . 27/8=9/16`
`4//|(-0,5)^2+7/2|.10-(29/30-7/15):(-2017/2018)^0=|1/4+7/2|.10-1/2:1=|15/4|.10-1/2=15/4 .10-1/2=75/2-1/2=37`
`5// 8/3+(3-1/2)^2-|[-7]/3|=8/3+(5/2)^2-7/3=8/3+25/4-7/3=107/12-7/3=79/12`
Cách 1 : Trước hết tính giá trị từng biểu thức trong ngoặc
\(-\left(4-\dfrac{1}{3}+\dfrac{3}{5}\right)-\left(3+\dfrac{2}{3}-\dfrac{4}{5}\right)+\left(5+\dfrac{4}{3}-\dfrac{4}{5}\right)\)
\(=-\left(\dfrac{60}{15}-\dfrac{5}{15}+\dfrac{9}{15}\right)-\left(\dfrac{45}{15}+\dfrac{10}{15}-\dfrac{20}{15}\right)+\left(\dfrac{75}{15}+\dfrac{20}{15}-\dfrac{12}{15}\right)\)
\(=-\left(\dfrac{64}{15}\right)-\left(\dfrac{35}{15}\right)+\left(\dfrac{83}{15}\right)\)
\(=-\dfrac{64}{15}-\dfrac{35}{15}+\dfrac{83}{15}\)
\(=-\dfrac{16}{15}\)
\(=-\dfrac{8}{5}\)
Cách 2 : Bỏ dấu ngoặc rồi nhóm các số hạng thích hợp
\(-\left(4-\dfrac{1}{3}+\dfrac{3}{5}\right)-\left(3+\dfrac{2}{3}-\dfrac{4}{5}\right)+\left(5+\dfrac{4}{3}-\dfrac{4}{5}\right)\)
\(=-4+\dfrac{1}{3}-\dfrac{3}{5}-3-\dfrac{2}{3}+\dfrac{4}{5}+5+\dfrac{4}{3}-\dfrac{4}{5}\)
\(=\left(-4-3+5\right)+\left(\dfrac{1}{3}-\dfrac{2}{3}+\dfrac{4}{3}\right)+\left(-\dfrac{3}{5}+\dfrac{4}{5}-\dfrac{4}{5}\right)\)
\(=-2+1-\dfrac{3}{5}\)
\(=-\dfrac{8}{5}\)
Cậu ơi, câu đầu tớ vẫn chưa hiểu ý câu hỏi là như nào hết, cậu có thể viết lại cho tớ đko ạ?
Câu 1 là : \(\dfrac{-2}{5}:1\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^2\)
Thực hiện các phép tính:
a) 9,6.212−(2.125−1512):149,6.212−(2.125−1512):14
b) 518−1,456:725+4,5.45518−1,456:725+4,5.45;
c) (12+0,8−113).(2,3+4725−1,28)(12+0,8−113).(2,3+4725−1,28)
d) (−5).12:[(−14)+12:(−2)]+113(−5).12:[(−14)+12:(−2)]+113.
Hướng dẫn làm bài:
a) 9,6.212−(2.125−1512):149,6.212−(2.125−1512):14
=9,6.52−(250−1712)×4=9,6.52−(250−1712)×4
=4,8.5−(1000−173)=4,8.5−(1000−173)
=24−1000+173=24−1000+173
=−976+173=−976+173
=−97013=−97013
b) 518−1,456:725+4,5.45518−1,456:725+4,5.45;
=518−1,456×257+92.45=518−1,456×257+92.45
=518−0,208×25+185=518−0,208×25+185
=518−5,2+185=518−5,2+185
=25−468+32490=25−468+32490
=−11990=−11990
c) (12+0,8−113).(2,3+4725−1,28)(12+0,8−113).(2,3+4725−1,28)
=(12+45−43).(2310+10725−3225)=(12+45−43).(2310+10725−3225)
=(15+24−4030).(2310+10725−3225)=(15+24−4030).(2310+10725−3225)
=(15+24−4030).(115+214−6450)=(15+24−4030).(115+214−6450)
=−130.26550=−130.26550
=−53300=−53300
d) (−5).12:[(−14)+12:(−2)]+113(−5).12:[(−14)+12:(−2)]+113
=−60:[14+12×(−12)]+1.13=−60:[14+12×(−12)]+1.13
=−60:[−14−14]+113=−60:[−14−14]+113
=−60:(12)+113=−60:(12)+113
=120+113=120+113
=12113
a) \(9,6.2\dfrac{1}{2}-\left(2.125-1\dfrac{5}{12}\right):\dfrac{1}{4}\)
\(=9,6.\dfrac{5}{2}-\left(250-\dfrac{17}{12}\right).4\)
\(=4,8.5-\left(1000-\dfrac{17}{3}\right)\)
\(=24-1000+\dfrac{17}{3}\)
\(=-976+\dfrac{17}{3}=-970\dfrac{1}{3}\)
b) \(\dfrac{5}{18}-1,456:\dfrac{7}{25}+4,5.\dfrac{4}{5}\)
\(=\dfrac{5}{18}-1,456.\dfrac{25}{7}+\dfrac{9}{2}.\dfrac{4}{5}\)
\(=\dfrac{5}{18}-0,208.25+\dfrac{18}{5}\)
\(=\dfrac{5}{18}-5,2+\dfrac{18}{5}\)
\(=-\dfrac{119}{90}\)
c) \(\left(\dfrac{1}{2}+0,8-1\dfrac{1}{3}\right).\left(2,3+4\dfrac{7}{25}-1,28\right)\)
\(=\left(\dfrac{1}{2}+\dfrac{4}{5}-\dfrac{4}{3}\right).\left(\dfrac{23}{10}+\dfrac{107}{25}-\dfrac{32}{25}\right)\)
\(=-\dfrac{1}{30}.\dfrac{265}{50}=-\dfrac{53}{300}\)
d) \(\left(-5\right).12:\left[\left(-\dfrac{1}{4}\right)+\dfrac{1}{2}:\left(-2\right)\right]+1\dfrac{1}{3}\)
\(=-60:\left[\dfrac{1}{4}+\dfrac{1}{2}.\dfrac{-1}{2}\right]+1.\dfrac{1}{3}\)
\(=-60:\left[-\dfrac{1}{4}-\dfrac{1}{4}\right]+1\dfrac{1}{3}\)
\(=-60:\left(\dfrac{1}{2}\right)+1\dfrac{1}{3}\)
\(=121\dfrac{1}{3}\)
A = (\(\dfrac{5}{6}\) - \(\dfrac{4}{5}\)) . 1\(\dfrac{1}{5}\) + \(\dfrac{3}{16}\) : (\(\dfrac{-1}{2}\))3
A = \(\dfrac{1}{30}\) . \(\dfrac{6}{5}\) + \(\dfrac{3}{16}\) : \(\dfrac{-1}{8}\)
A = \(\dfrac{1}{25}\) + \(\dfrac{3}{16}\) . \(\dfrac{-8}{1}\)
A = \(\dfrac{1}{25}\) + \(\dfrac{-3}{2}\)
A = \(\dfrac{-73}{50}\)
B = \(\dfrac{4}{17}\) . (7\(\dfrac{3}{4}\) - 6\(\dfrac{1}{3}\)) + (5\(\dfrac{3}{4}\) - 6.95) : (-1\(\dfrac{3}{5}\))
B = \(\dfrac{4}{17}\) . \(\dfrac{17}{12}\) + (\(\dfrac{23}{4}\) - \(\dfrac{139}{20}\)) : \(\dfrac{-8}{5}\)
B = \(\dfrac{1}{3}\) + \(\dfrac{-6}{5}\) . \(\dfrac{-5}{8}\)
B = \(\dfrac{13}{12}\)
=\(\dfrac{3}{2}+\dfrac{1}{5}-\dfrac{3}{4}\)
=\(\dfrac{30}{20}+\dfrac{4}{20}-\dfrac{15}{20}\)
=\(\dfrac{19}{20}\)
=\(\dfrac{4}{3}+\dfrac{2}{3}-\dfrac{3}{4}-\dfrac{4}{5}-\dfrac{6}{5}\)
=(\(\dfrac{4}{3}+\dfrac{2}{3}\)) -\(\dfrac{4}{3}-\left(\dfrac{4}{5}-\dfrac{6}{5}\right)\)
=2\(-\dfrac{4}{3}-\dfrac{-2}{5}\)
=\(\dfrac{30}{15}-\dfrac{20}{15}-\dfrac{-6}{15}\)
=\(\dfrac{4}{15}\)