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a: \(A=49+\dfrac{8}{23}-14-\dfrac{8}{23}-5-\dfrac{7}{32}=30-\dfrac{7}{32}=\dfrac{953}{32}\)
b:
Sửa đề: \(B=71\dfrac{38}{45}-\left(43\dfrac{8}{45}-1\dfrac{17}{51}\right)\)
\(B=71+\dfrac{38}{45}-43-\dfrac{8}{45}+1+\dfrac{17}{51}\)
\(=71-43+1+1\)
=28+2=30
1: \(=\dfrac{15}{37}\cdot\dfrac{38}{41}-\dfrac{15}{37}\cdot\dfrac{74}{45}-\dfrac{38}{41}\cdot\dfrac{15}{37}-\dfrac{38}{41}\cdot\dfrac{82}{76}\)
\(=\dfrac{-2}{3}-1=-\dfrac{5}{3}\)
2: \(=\dfrac{47}{53}\cdot\dfrac{17}{3}-\dfrac{47}{53}\cdot\dfrac{53}{47}+\dfrac{17}{3}\cdot\dfrac{6}{17}-\dfrac{17}{3}\cdot\dfrac{47}{53}\)
\(=-1+2=1\)
Trời ơi cái đề bài !!!
Thoy thì làm từng câu vậy
a) \(I=10101.\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{111111}\right)\)
\(I=10101.\left(\dfrac{10}{222222}+\dfrac{5}{222222}-\dfrac{8}{222222}\right)\)
\(I=10101.\left(\dfrac{15}{222222}-\dfrac{8}{222222}\right)\)
\(I=10101.\dfrac{7}{222222}\)
\(I=\dfrac{7}{22}\)
Mấy bài này bạn tự làm đi, chuyển vế tìm x gần giống cấp I mà.
b)\(\dfrac{-3}{5}.x=\dfrac{1}{4}+0,75\)
=>\(\dfrac{-3}{5}.x=1\)
=>\(x=1:\dfrac{-3}{5}\)
=>\(x=\dfrac{-5}{3}\)
Vậy \(x=\dfrac{-5}{3}\)
13)\(\dfrac{45}{8}\left(\dfrac{4}{15}-\dfrac{7}{8}\right)+\dfrac{45}{8}\left(\dfrac{11}{15}+\dfrac{9}{8}\right)\)
=\(\dfrac{45}{8}\left(\dfrac{4}{15}-\dfrac{7}{8}+\dfrac{11}{15}+\dfrac{9}{8}\right)\)
=\(\dfrac{45}{8}\left[\left(\dfrac{4}{15}+\dfrac{11}{15}\right)-\left(\dfrac{7}{8}-\dfrac{9}{8}\right)\right]\)
=\(\dfrac{45}{8}.\dfrac{5}{4}\)=\(\dfrac{225}{32}\)
14)\(\dfrac{15}{7}\left(\dfrac{49}{35}-\dfrac{27}{8}\right)-\left(\dfrac{2}{5}-\dfrac{27}{4}\right):\dfrac{7}{15}\)
=\(\dfrac{15}{7}\left(\dfrac{49}{35}-\dfrac{27}{8}\right)-\left(\dfrac{2}{5}-\dfrac{27}{4}\right).\dfrac{15}{7}\)
=\(\dfrac{15}{7}\left[\left(\dfrac{49}{35}-\dfrac{27}{8}\right)-\left(\dfrac{2}{5}-\dfrac{27}{4}\right)\right]\)
=\(\dfrac{15}{7}\left(\dfrac{49}{35}-\dfrac{27}{8}-\dfrac{2}{5}+\dfrac{27}{4}\right)\)
=\(\dfrac{15}{7}.\dfrac{35}{8}\)=\(\dfrac{75}{8}\)
Bài 1:
a: \(\Leftrightarrow\dfrac{2}{3}\cdot\dfrac{6+9-4}{12}< =\dfrac{x}{18}< =\dfrac{7}{13}\cdot\dfrac{3-1}{6}\)
\(\Leftrightarrow\dfrac{22}{36}< =\dfrac{x}{18}< =\dfrac{14}{78}=\dfrac{7}{39}\)
\(\Leftrightarrow\dfrac{11}{9}< =\dfrac{x}{9}< =\dfrac{7}{13}\)
=>143<=x<=63
hay \(x\in\varnothing\)
b: \(\Leftrightarrow\dfrac{31\cdot9-26\cdot4}{180}\cdot\dfrac{-36}{35}< x< \dfrac{153+64+56}{168}\cdot\dfrac{8}{13}\)
\(\Leftrightarrow-1< x< 1\)
=>x=0
3)\(\dfrac{-41}{32}\left(\dfrac{15}{8}-\dfrac{16}{41}\right)+\dfrac{15}{8}\left(\dfrac{41}{32}-\dfrac{8}{3}\right)\)
=\(\dfrac{-41}{32}.\dfrac{15}{8}-\dfrac{-41}{32}.\dfrac{16}{41}+\dfrac{15}{8}.\dfrac{41}{32}-\dfrac{15}{8}.\dfrac{8}{3}\)
=\(\left(\dfrac{-41}{32}.\dfrac{15}{8}+\dfrac{15}{8}.\dfrac{41}{32}\right)+\dfrac{-16}{41}.\dfrac{-41}{32}-\dfrac{15}{8}.\dfrac{8}{3}\)
=\(0+\dfrac{1}{2}-5=\dfrac{-9}{2}\)
4)\(\dfrac{13}{29}\left(\dfrac{29}{5}-\dfrac{45}{8}\right)-\dfrac{45}{8}\left(\dfrac{9}{8}-\dfrac{13}{29}\right)\)
=\(\dfrac{13}{29}.\dfrac{29}{5}-\dfrac{45}{8}.\dfrac{13}{29}-\dfrac{45}{8}.\dfrac{9}{8}-\dfrac{45}{8}.\dfrac{13}{29}\)
=\(\left(\dfrac{45}{8}.\dfrac{13}{29}-\dfrac{45}{8}.\dfrac{13}{29}\right)-\dfrac{13}{29}.\dfrac{29}{5}-\dfrac{45}{8}.\dfrac{9}{8}\)
=\(0-\dfrac{13}{5}-\dfrac{405}{64}=\dfrac{-2857}{320}\)
\(\dfrac{38}{45}-\left(\dfrac{8}{45}-\dfrac{17}{51}-\dfrac{3}{11}\right)\)
\(=\dfrac{38}{45}-\dfrac{8}{45}+\dfrac{17}{51}+\dfrac{3}{11}\)
\(=\dfrac{2}{3}+\dfrac{1}{3}+\dfrac{3}{11}\)
\(=1\dfrac{3}{11}\)