Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a,(3/5+0,415-3/200).\(2\dfrac{2}{3}\).0,25
=(0,6+0.415-3/200)8/3.1/4
=(1,015-3/200).8/3.1/4
=(1015/1000-3/200).2/3
=(203/200-3/200).2/3
=1.2/3=2/3
b,0,25:(10,3-9,8)-3/4
=0,25:910,3-9,8)-0.75
=0,25:0,5-0,75
=-0,25
Mình làm từng đó trước,lần sau mình sẽ lm típ nha!
Mình tick đúng nhưng bạn cần đọc rõ đề Tính theo cách hợp lý nhất
Bài 1:
a) \(\left(\dfrac{3}{8}+\dfrac{-3}{4}+\dfrac{7}{12}\right):\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\left(\dfrac{9}{24}+\dfrac{-18}{24}+\dfrac{14}{24}\right):\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\dfrac{5}{24}:\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\dfrac{5}{24}.\dfrac{6}{5}+\dfrac{1}{2}\)
\(=\dfrac{1}{4}+\dfrac{1}{2}\)
\(=\dfrac{1}{4}+\dfrac{2}{4}\)
\(=\dfrac{3}{4}\)
b) \(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\)
\(=\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\)
\(=\left(\dfrac{1}{2}+\dfrac{4}{5}\right)+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)\)
\(=\dfrac{1}{2}+\dfrac{4}{5}\)
\(=\dfrac{5}{10}+\dfrac{8}{10}\)
\(=\dfrac{9}{5}\)
c) \(6\dfrac{5}{12}:2\dfrac{3}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{77}{12}:\dfrac{11}{4}+\dfrac{42}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{77}{12}.\dfrac{4}{11}+\dfrac{42}{4}.\left(\dfrac{5}{15}+\dfrac{3}{15}\right)\)
\(=\dfrac{7}{3}+\dfrac{42}{4}.\dfrac{8}{15}\)
\(=\dfrac{7}{3}+\dfrac{14.2}{1.3}\)
\(=\dfrac{7}{3}+\dfrac{28}{3}\)
\(=\dfrac{35}{3}\)
d) \(\left(\dfrac{7}{8}-\dfrac{3}{4}\right).1\dfrac{1}{3}-\dfrac{2}{7}.\left(3,5\right)^2\)
\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right).\dfrac{4}{3}-\dfrac{2}{7}.12\dfrac{1}{4}\)
\(=\dfrac{1}{8}.\dfrac{4}{3}-\dfrac{2}{7}.\dfrac{49}{4}\)
\(=\dfrac{1}{6}-\dfrac{7}{2}\)
\(=\dfrac{1}{6}-\dfrac{21}{6}\)
\(=\dfrac{-10}{3}\)
e) \(\left(\dfrac{3}{5}+0,415-\dfrac{3}{200}\right).2\dfrac{2}{3}.0,25\)
\(=\left(\dfrac{3}{5}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)
\(=\left(\dfrac{120}{200}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)
\(=1.\dfrac{8}{3}.\dfrac{1}{4}\)
\(=\dfrac{2}{3}\)
f) \(\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)
\(=\dfrac{5}{16}:\dfrac{1}{8}-\left(\dfrac{9}{4}-\dfrac{3}{5}\right).\dfrac{10}{11}\)
\(=\dfrac{5}{16}.\dfrac{8}{1}-\left(\dfrac{45}{20}-\dfrac{12}{20}\right).\dfrac{10}{11}\)
\(=\dfrac{5}{2}-\dfrac{33}{20}.\dfrac{10}{11}\)
\(=\dfrac{5}{2}-\dfrac{3}{2}\)
\(=\dfrac{2}{2}=1\)
g) \(0,25:\left(10,3-9,8\right)-\dfrac{3}{4}\)
\(=\dfrac{1}{4}:\dfrac{1}{2}-\dfrac{3}{4}\)
\(=\dfrac{1}{4}.\dfrac{2}{1}-\dfrac{3}{4}\)
\(=\dfrac{1}{2}-\dfrac{3}{4}\)
\(=\dfrac{2}{4}-\dfrac{3}{4}\)
\(=\dfrac{-1}{4}\)
h) \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+20\%\right):\dfrac{7}{3}\)
\(=\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{1}{5}\right):\dfrac{7}{3}\)
\(=\dfrac{7}{5}-\left(\dfrac{11}{20}+\dfrac{4}{20}\right):\dfrac{7}{3}\)
\(=\dfrac{7}{5}-\dfrac{3}{4}:\dfrac{7}{3}\)
\(=\dfrac{7}{5}-\dfrac{9}{28}\)
\(=\dfrac{196}{140}-\dfrac{45}{140}\)
\(=\dfrac{151}{140}\)
i) \(\dfrac{\left(\dfrac{1}{2-0,75}\right).\left(0,2-\dfrac{2}{5}\right)}{\dfrac{5}{9}-1\dfrac{1}{12}}\)
\(=\dfrac{\left(\dfrac{1}{1,25}\right).\left(\dfrac{1}{5}-\dfrac{2}{5}\right)}{\dfrac{5}{9}-\dfrac{13}{12}}\)
\(=\dfrac{\dfrac{1}{1,25}.\dfrac{-1}{5}}{\dfrac{20}{36}-\dfrac{39}{36}}\)
\(=\dfrac{\dfrac{-1}{6,25}}{\dfrac{-19}{36}}\)
k) \(\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{1}{14}}{-1-\dfrac{3}{7}+\dfrac{3}{28}}\)
\(=\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{28}}{-\dfrac{3}{3}-\dfrac{3}{7}+\dfrac{3}{28}}\)
\(=\dfrac{2\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}{\left(-3\right)\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}\)
\(=-\dfrac{2}{3}\)
\(A=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(A=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)
\(A=\left(\dfrac{7}{10}.\dfrac{5}{28}\right).\left(\dfrac{8}{3}.\dfrac{3}{8}\right).20\)
\(A=\dfrac{1}{8}.1.20\)
\(A=\dfrac{20}{8}=\dfrac{5}{2}\)
\(B=\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)
\(B=\left(9\dfrac{3}{8}+7\dfrac{5}{8}\right)+4,03\)
\(B=\left[\left(9+7\right)+\left(\dfrac{3}{8}+\dfrac{5}{8}\right)\right]+4,03\)
\(B=\left(16+1\right)+4,03\)
\(B=17+4,03\)
\(B=21,03\)
\(C=\left(9,75.21\dfrac{3}{7}+\dfrac{39}{4}.18\dfrac{4}{7}\right).\dfrac{15}{78}\)
\(C=\left(\dfrac{39}{4}.\dfrac{150}{7}+\dfrac{39}{4}.\dfrac{130}{7}\right).\dfrac{15}{78}\)
\(C=\dfrac{39}{4}.\left(\dfrac{150}{7}+\dfrac{130}{7}\right).\dfrac{15}{78}\)
\(C=\dfrac{39}{4}.40.\dfrac{15}{78}\)
\(C=390.\dfrac{15}{78}\)
\(C=75\)
Bài làm
a) \(\left(\frac{3}{8}+\frac{-1}{4}+\frac{5}{12}\right):\frac{2}{3}\)
\(=\left(\frac{9}{24}+\frac{-6}{24}+\frac{10}{24}\right).\frac{3}{2}\)
\(=\frac{13}{24}.\frac{3}{2}\)
\(=\frac{39}{48}\)
b) \(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+1\frac{5}{7}\)
\(=-\frac{5}{7}.\left(\frac{2}{11}+\frac{9}{11}\right)+\frac{12}{7}\)
\(=-\frac{5}{7}.1+\frac{12}{7}\)
\(=\frac{7}{7}\)
\(=1\)
c) \(0,25:\left(10,3-9,8\right)-\frac{3}{4}\)
\(=\frac{25}{100}:\left(0,5\right)-\frac{3}{4}\)
\(=\frac{5}{4}:\frac{1}{2}-\frac{3}{4}\)
\(=\frac{5}{4}.2-\frac{3}{4}\)
\(=\frac{5}{2}-\frac{3}{4}\)
\(=\frac{10}{4}-\frac{3}{4}=\frac{7}{4}\)
d) \(-\frac{5}{9}.\frac{13}{28}-\frac{13}{28}.\frac{4}{9}\)
\(=-\frac{13}{28}\left(\frac{5}{9}+\frac{4}{9}\right)\)
\(=-\frac{13}{28}.\frac{9}{9}\)
\(=-\frac{13}{28}.1\)
\(=-\frac{13}{28}\)
\(1,A=-\dfrac{3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)
\(A=-\dfrac{3}{4}.\left(0,125-\dfrac{3}{2}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=-\dfrac{3}{4}.\left(-\dfrac{11}{8}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}:\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}.\dfrac{16}{33}-\dfrac{1}{4}\)
\(A=\dfrac{1}{2}-\dfrac{1}{4}\)
\(A=\dfrac{2}{4}-\dfrac{1}{4}\)
\(A=\dfrac{1}{4}\)
\(a,\frac{5}{16}:0,125-\left(2\frac{1}{4}-0,6\right)\cdot\frac{10}{11}\)
\(=\frac{5}{16}:\frac{1}{8}-\left(\frac{9}{4}-\frac{3}{5}\right)\cdot\frac{10}{11}\)
\(=\frac{5}{2}-\frac{33}{20}\cdot\frac{10}{11}\)
\(=\frac{5}{2}-\frac{3}{2}\)
\(=1\)
\(\frac{5}{16}:0,125-\left(2\frac{1}{4}-0,6\right).\frac{10}{11}\)
\(\Rightarrow\frac{5}{16}:\frac{125}{1000}-\left(\frac{9}{4}-\frac{6}{10}\right).\frac{10}{11}\)
\(\Rightarrow\frac{5}{16}:\frac{1}{8}-\left(\frac{9}{4}-\frac{3}{5}\right).\frac{10}{11}\)
\(\Rightarrow\frac{5}{16}.\frac{8}{1}-\left(\frac{45}{20}-\frac{12}{20}\right).\frac{10}{11}\)
\(\Rightarrow\frac{5}{2}-\frac{33}{20}.\frac{10}{11}\)
\(\Rightarrow\frac{5}{2}-\frac{3.1}{2.1}\)
\(\Rightarrow\frac{5}{2}-\frac{3}{2}=\frac{2}{2}=1\)
a=113/13-(24/7+53/13)=113/13-24/7-53/13=60/13-24/7=396/16
b=(64/9+37/11)-44/9=64/9+37/11-44/9=20/9+37/11=181/33
c=-5/7(2/11+9/11)+15/7=-5/7+15/7=10/7
d=0.7.22/3.20.0.375.5/28=0
e=(6,17+35/9-236/97)(1/3-0.25-1/12)=A.(1/3-1/4-1/12)=A.(1/12-1/12)=A.0=0
e=(6,17 + 3 5/9 - 2 36/97 ) . ( 1/3 - 0,25 - 1/12)=(6,17 + 3 5/9 - 2 36/97 ) .(1/3-1/12-0,25)
=(6,17 + 3 5/9 - 2 36/97 ) .(4/12-1/12-0,25)=(6,17 + 3 5/9 - 2 36/97 ) .(3/12-0,25)
=(6,17 + 3 5/9 - 2 36/97 ) .(1/4-0.25)=(6,17 + 3 5/9 - 2 36/97 ) .(0.25-0.25)
=(6,17 + 3 5/9 - 2 36/97 ) .0=0
Vậy E=0
a) \(\left(\frac{7}{8}-\frac{3}{4}\right)\cdot\frac{1}{3}-\frac{2}{7}\cdot\left(3,5\right)=\left(\frac{7}{8}-\frac{3}{4}\right)\cdot\frac{1}{3}-\frac{2}{7}\cdot\frac{7}{2}\)
\(=\left(\frac{7}{8}-\frac{6}{8}\right)\cdot\frac{1}{3}-1=\frac{1}{8}\cdot\frac{1}{3}-1=\frac{1}{24}-\frac{24}{24}=-\frac{23}{24}\)
b) \(\left(\frac{3}{5}+0,415-\frac{3}{200}\right)\cdot2\frac{2}{3}\cdot0,25\)
\(=\left(\frac{3}{5}+\frac{83}{200}-\frac{3}{200}\right)\cdot\frac{8}{3}\cdot\frac{1}{4}\)
\(=\left(\frac{3}{5}+\frac{80}{200}\right)\cdot\frac{8}{3}\cdot\frac{1}{4}=\left(\frac{3}{5}+\frac{2}{5}\right)\cdot\frac{8}{3}\cdot\frac{1}{4}=1\cdot\frac{8}{3}\cdot\frac{1}{4}=1\cdot\frac{2}{3}\cdot\frac{1}{1}=\frac{2}{3}\)
c) \(\frac{5}{16}:0,125-\left(2\frac{1}{4}-0,6\right)\cdot\frac{10}{11}\)
\(=\frac{5}{16}:\frac{1}{8}-\left(\frac{9}{4}-\frac{3}{5}\right)\cdot\frac{10}{11}\)
\(=\frac{5}{16}\cdot8-\frac{33}{20}\cdot\frac{10}{11}=\frac{5}{2}-\frac{3}{2}=1\)
d) \(0,25:\left(10,3-9,8\right)-\frac{3}{4}=\frac{1}{4}:\left(\frac{103}{10}-\frac{98}{10}\right)-\frac{3}{4}\)
\(=\frac{1}{4}:\frac{1}{2}-\frac{3}{4}=\frac{1}{4}\cdot2-\frac{3}{4}=\frac{2}{4}-\frac{3}{4}=-\frac{1}{4}\)
Câu cuối tương tự