a) Ta có: \(2\dfrac{3}{3}\cdot4\cdot\left(-0.4\right)+1\dfrac{3}{5}\cdot1.75+\left(-7.2\right):\dfrac{9}{11}\)

\(=-4.8+\dfrac{8}{5}\cdot\dfrac{7}{4}-\dfrac{36}{5}\cdot\dfrac{11}{9}\)

\(=\dfrac{-24}{5}+\dfrac{14}{5}-\dfrac{44}{5}\)

\(=\dfrac{-54}{5}\)

b) Ta có: \(\left(\dfrac{1}{24}-\dfrac{5}{16}\right):\dfrac{-3}{8}+1^{10}\cdot\left(-5\right)^0\)

\(=\left(\dfrac{2}{48}-\dfrac{15}{48}\right)\cdot\dfrac{8}{-3}+1\cdot1\)

\(=\dfrac{-13}{48}\cdot\dfrac{-8}{3}+1\)

\(=\dfrac{13}{18}+\dfrac{18}{18}=\dfrac{31}{18}\)

Kết quả học tập kì I của lớp 6A xếp thành 3 loại:giỏi,khá,trung bình.Số học sinh giỏi chiếm \(\dfrac{1}{3}\) số học sinh của lớp,số học sinh khá chiếm 40% số học sinh cả lớp,số học sinh trung bình là 12 em.Tính số học sinh lớp 6A và tỉ số phần trăm của học sinh giỏi so với học sinh cả lớp.

a: \(0.4\cdot\sqrt{0.25-\sqrt{\dfrac{1}{4}}}=0.4\cdot\sqrt{0.25-0.5}\)(đề này sai rồi bạn)

b: \(\dfrac{3}{2}+2\left(x-1\right)=-5\dfrac{1}{2}\)

\(\Leftrightarrow2\left(x-1\right)=\dfrac{-11}{2}-\dfrac{3}{2}=-7\)

\(\Leftrightarrow x-1=\dfrac{-7}{2}\)

hay \(x=-\dfrac{5}{2}\)

\(a,\left(7+3\dfrac{1}{4}-\dfrac{3}{5}\right)+\left(0,4-5\right)-\left(4\dfrac{1}{4}-1\right)\)

\(=\left(7+\dfrac{13}{4}-\dfrac{3}{5}\right)-\dfrac{23}{5}-\left(\dfrac{17}{4}-1\right)\)

\(=7+\dfrac{13}{4}-\dfrac{3}{5}-\dfrac{23}{5}-\dfrac{17}{4}+1\)

\(=\left(7+1\right)+\left(\dfrac{13}{4}-\dfrac{17}{4}\right)-\left(\dfrac{3}{5}+\dfrac{23}{5}\right)\)

\(=8-\dfrac{4}{4}-\dfrac{26}{5}\)

\(=7-\dfrac{26}{5}\)

\(=\dfrac{9}{5}\)

\(b,\dfrac{2}{3}-\left[\left(-\dfrac{7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)

\(=\dfrac{2}{3}-\left(-\dfrac{7}{4}-\dfrac{1}{2}-\dfrac{3}{8}\right)\)

\(=\dfrac{2}{3}-\left(-\dfrac{14}{8}-\dfrac{4}{8}-\dfrac{3}{8}\right)\)

\(=\dfrac{2}{3}-\left(-\dfrac{21}{8}\right)\)

\(=\dfrac{2}{3}+\dfrac{21}{8}\)

\(=\dfrac{79}{24}\)

\(c,\left(9-\dfrac{1}{2}-\dfrac{3}{4}\right):\left(7-\dfrac{1}{4}-\dfrac{5}{8}\right)\)

\(=\left(\dfrac{36}{4}-\dfrac{2}{4}-\dfrac{3}{4}\right):\left(\dfrac{56}{8}-\dfrac{2}{8}-\dfrac{5}{8}\right)\)

\(=\dfrac{31}{4}:\dfrac{49}{8}\)

\(=\dfrac{62}{49}\)

\(d,3-\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}}=3-\dfrac{\dfrac{7}{7}-\dfrac{1}{7}}{\dfrac{7}{7}+\dfrac{1}{7}}=3-\left(\dfrac{6}{7}:\dfrac{8}{7}\right)=3-\dfrac{3}{4}=\dfrac{9}{4}\)

\(a,0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{5}{7}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\dfrac{5}{6}+\dfrac{39}{35}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\left(\dfrac{5}{6}-\dfrac{1}{6}\right)+\left(\dfrac{39}{35}-\dfrac{4}{35}\right)\\ =\dfrac{2}{3}+1\\ =\dfrac{4}{3}.\)

\(b,\left(3-\dfrac{1}{4}+\dfrac{2}{3}\right)-\left(5+\dfrac{1}{3}-\dfrac{6}{5}\right)-\left(-6-\dfrac{7}{4}+\dfrac{3}{2}\right)\\ =3-\dfrac{1}{4}+\dfrac{2}{3}-5-\dfrac{1}{3}+\dfrac{6}{5}+6+\dfrac{7}{4}-\dfrac{3}{2}\\ =\left(3-5+6\right)+\left(-\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{2}{3}-\dfrac{1}{3}\right)+\left(\dfrac{6}{5}+\dfrac{7}{4}\right)\\ =4-\dfrac{3}{2}+\dfrac{1}{3}+\dfrac{59}{20}\\ =\dfrac{5}{2}+\dfrac{1}{3}+\dfrac{59}{20}\\ =\dfrac{17}{6}+\dfrac{59}{20}\\ =\dfrac{347}{60}.\)

\(c,\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\\ =\dfrac{1}{3}+\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\\ =\left(\dfrac{1}{3}-\dfrac{2}{9}\right)+\left(\dfrac{3}{4}-\dfrac{1}{36}\right)+\left(\dfrac{3}{5}+\dfrac{1}{15}\right)+\dfrac{1}{64}\\ =\dfrac{1}{9}+\dfrac{13}{18}+\dfrac{2}{3}+\dfrac{1}{64}\\ =\dfrac{3}{2}+\dfrac{1}{64}\\ =\dfrac{65}{64}.\)

Cái này bn lầy máy tính ra tính tí là xong thôi

a) \(2\dfrac{3}{4}.\left(-0,4\right)-1\dfrac{3}{5}.2,75+\left(-1,2\right):\dfrac{4}{11}\)

= \(2,75.\left(-0,4\right)-\left(1,6\right).\left(2,75\right)+\left(-1,2\right).\dfrac{11}{4}\)

= \(2,75.\left(-0,4\right)-\left(1,6\right).\left(2,75\right)+\left(-1,2\right).\left(2,75\right)\)

= \(2,75.\left\{\left(-0,4\right)-\left(1,6\right)+\left(-1,2\right)\right\}\)

= \(2,75.\left(-3,2\right)\)

= \(-8,8\)

b) \(1,4.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):2\dfrac{1}{5}\)

= \(\dfrac{7}{5}.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):\dfrac{11}{5}\)

= \(\dfrac{7}{5}.\dfrac{15}{49}-\dfrac{22}{15}.\dfrac{5}{11}\)

= \(\dfrac{3}{7}-\dfrac{2}{3}\)

= \(-\dfrac{5}{21}\)

c) \(\left(-3,2\right).\dfrac{15}{64}+\left(0,8-2\dfrac{4}{15}\right):3\dfrac{2}{3}\)

= \(-\dfrac{16}{5}.\dfrac{15}{64}+\left(\dfrac{4}{5}-2\dfrac{4}{15}\right):\dfrac{11}{3}\)

= \(-\dfrac{16}{5}.\dfrac{15}{64}+\left(-\dfrac{22}{15}\right).\dfrac{3}{11}\)

= \(\left(-\dfrac{3}{4}\right)+\left(-\dfrac{2}{5}\right)\)

= \(-\dfrac{23}{20}\)

d) \(0,02.\dfrac{-25}{2}+\dfrac{3}{8}+\left(-2\dfrac{9}{20}\right).\dfrac{2}{7}\)

= \(\dfrac{1}{50}.\dfrac{-25}{2}+\dfrac{3}{8}+\left(-\dfrac{49}{20}\right).\dfrac{2}{7}\)

=\(\left(-\dfrac{1}{4}\right)+\dfrac{3}{8}+\left(-\dfrac{7}{10}\right)\)

= \(\dfrac{1}{8}+\left(-\dfrac{7}{10}=\right)\)

= \(-\dfrac{23}{40}\)

e) \(34\%:\dfrac{51}{16}-3\dfrac{7}{9}.6,5-\left(0,4\right)^2\)

= \(\dfrac{17}{50}.\dfrac{16}{51}-\dfrac{34}{9}.\dfrac{13}{2}-\dfrac{4}{25}\)

= \(\dfrac{8}{75}-\dfrac{221}{9}-\dfrac{4}{15}\)

= \(-\dfrac{5501}{225}\)

\(A=\left(3-\dfrac{1}{4}+\dfrac{3}{2}\right)-\left(5+\dfrac{1}{3}-\dfrac{5}{6}\right)-\left(6-\dfrac{7}{4}+\dfrac{2}{3}\right)\\ \Rightarrow A=3-\dfrac{1}{4}+\dfrac{3}{2}-5-\dfrac{1}{3}+\dfrac{5}{6}-6+\dfrac{7}{4}-\dfrac{2}{3}\\ \Rightarrow A=\left(3-5-6\right)-\left(\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{3}{2}+\dfrac{5}{6}-\dfrac{2}{3}\right)\\ \Rightarrow A=-8-\dfrac{3}{2}+\dfrac{5}{3}\\ =-\dfrac{47}{6}.\\ B=0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{41}\)

\(\Rightarrow B=\left(0,5+0,4\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{5}{7}-\dfrac{4}{35}\right)+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{9}{10}+\dfrac{1}{2}+\dfrac{3}{5}+\dfrac{1}{41}\\ \Rightarrow B=2+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{83}{41}.\)

-_-

\(c,2\dfrac{1}{2}.\left(\dfrac{5}{2}-30\%\right)+2\\ =\dfrac{5}{2}.\left(\dfrac{5}{2}-\dfrac{3}{10}\right)+2\\ =\dfrac{5}{2}.\left(\dfrac{5.5-3}{10}\right)+2\\ =\dfrac{5}{2}.\dfrac{11}{5}+2\\ =\dfrac{11}{2}+2\\ =\dfrac{11+2.2}{2}\\ =\dfrac{15}{2}\)

\(d,3\dfrac{1}{2}-0,4.\left(50\%-\dfrac{3}{2}\right)+\dfrac{3}{4}\\ =\dfrac{7}{2}-\dfrac{2}{5}.\left(\dfrac{1}{2}-\dfrac{3}{2}\right)+\dfrac{3}{4}\\ =\dfrac{7}{2}-\dfrac{2}{5}.\left(\dfrac{-2}{2}\right)+\dfrac{3}{4}\\ =\dfrac{7}{2}-\dfrac{2}{5}.\left(-1\right)+\dfrac{3}{4}\\ =\dfrac{7}{2}+\dfrac{2}{5}+\dfrac{3}{4}\\ =\dfrac{70}{20}+\dfrac{8}{20}+\dfrac{15}{20}\\ =\dfrac{93}{20}\)

qq

\(-2\dfrac{3}{4}.\left(-0,4\right)+1\dfrac{3}{5}.2,75-1,2:\dfrac{4}{11}\)

\(=\dfrac{11}{4}.\dfrac{-2}{5}+\dfrac{8}{5}.\dfrac{11}{4}-\dfrac{6}{5}:\dfrac{4}{11}\)

\(=\dfrac{11}{4}.\dfrac{-2}{5}+\dfrac{8}{5}.\dfrac{11}{4}-\dfrac{6}{5}.\dfrac{11}{4}\)

\(=\dfrac{11}{4}\left(\dfrac{-2}{5}+\dfrac{8}{5}-\dfrac{6}{5}\right)\)

\(=\dfrac{11}{4}.0\)

\(=0\)

\(\dfrac{2}{5}\times\dfrac{3}{4}-\dfrac{1}{8}=\dfrac{1}{5}\times\dfrac{3}{2}-\dfrac{1}{8}=\dfrac{3}{10}-\dfrac{1}{8}=\dfrac{24}{80}-\dfrac{10}{80}=\dfrac{14}{80}=\dfrac{7}{40}\\ \dfrac{4}{3}+\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{5}{3}-\dfrac{1}{5}=\dfrac{25}{15}-\dfrac{3}{15}=\dfrac{22}{15}\\ \dfrac{9}{20}-\dfrac{3}{5}\times\dfrac{1}{4}=\dfrac{9}{20}-\dfrac{3}{20}=\dfrac{6}{20}=\dfrac{3}{10}\\ \dfrac{2}{8}+\dfrac{2}{3}:\dfrac{4}{5}=\dfrac{2}{8}+\dfrac{2}{3}\times\dfrac{5}{4}=\dfrac{2}{8}+\dfrac{1}{3}\times\dfrac{5}{2}=\dfrac{2}{8}+\dfrac{5}{6}=\dfrac{1}{4}+\dfrac{5}{6}=\dfrac{6}{24}+\dfrac{20}{24}=\dfrac{26}{24}=\dfrac{13}{12}\)

   ( 2 + \(\dfrac{1}{3}\) - 0,4 ) - ( 7 - \(\dfrac{3}{5}\) - \(\dfrac{4}{3}\)) - ( \(\dfrac{1}{5}\) + \(\dfrac{5}{3}\)- 4)

=  2 + \(\dfrac{1}{3}\)- \(\dfrac{2}{5}\) - 7 + \(\dfrac{3}{5}\) + \(\dfrac{4}{3}\) - \(\dfrac{1}{5}\)  - \(\dfrac{5}{3}\) + 4

= ( 2 - 7 + 4) + ( \(\dfrac{1}{3}\) + \(\dfrac{4}{3}\)- \(\dfrac{5}{3}\)) + ( -\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\) - \(\dfrac{1}{5}\))

= -1 + 0 + 0

= - 1  

=(2+0,3-0,4)-(7-0,6-1,3)-(0,2+1,6)

=(2,3-0,4)-(7,6-1,3)-1,8

=1,9-6,3-1,8

= -2,6

=(2+0,3-0,4)-(7-0,6-1,3)-(0,2+1,6)

=(2,3-0,4)-(7,6-1,3)-1,8

=1,9-6,3-1,8

= -2,6