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\(\frac{13}{25}+\frac{6}{41}-\frac{38}{25}+\frac{35}{41}-\frac{1}{2}\)
\(=\left(\frac{13}{25}-\frac{38}{25}\right)+\left(\frac{6}{41}+\frac{35}{41}\right)-\frac{1}{2}\)
\(=-1+1-\frac{1}{2}=0-\frac{1}{2}\)
\(=\frac{-1}{2}\)
\(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
\(=\left(1\frac{4}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+0,5\)
\(=1+1+0,5=2,5\)
\(\frac{13}{25}+\frac{4}{41}-\frac{38}{25}+\frac{35}{41}-\frac{1}{2}\)
= \(\left(\frac{13}{25}-\frac{38}{25}\right)+\left(\frac{6}{41}+\frac{35}{41}\right)-\frac{1}{2}\)
= \(-1+1-\frac{1}{2}=-\frac{1}{2}\)
\(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
=\(\left(1\frac{4}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+0,5\)
= \(1+1+0,5=2,5\)
A) \(\frac{2}{3}+\frac{1}{5}-\frac{10}{7}=\frac{13}{15}-\frac{10}{7}=\frac{-59}{105}\)
B) \(\frac{7}{12}-\frac{27}{18}.\frac{2}{18}=\frac{7}{12}-\frac{3}{2}.\frac{2}{18}=\frac{7}{12}-\frac{1}{6}=\frac{5}{12}\)
C) \(\left(\frac{23}{41}-\frac{15}{82}\right).\frac{41}{25}=\frac{31}{82}.\frac{41}{25}=\frac{31}{50}\)
D) \(\left(\frac{4}{5}+\frac{1}{2}\right).\left(\frac{3}{13}-\frac{8}{13}\right)=\frac{13}{10}.\frac{-5}{13}=\frac{-1}{2}\)
a, = \(\frac{-59}{105}\)
b = \(\frac{7}{144}\)
c= \(\frac{31}{50}\)
d=\(\frac{-1}{2}\)
Trả lời
b)(1/3+12/67+13/41)-(79/67-28/41)
=1/3+12/67+13/41-79/67+28/41
=1/3+(12/67-79/67)+(13/41+28/41)
=1/3+(-67/67)+41/41
=1/3+(-1)+1
=1/3+0
=1/3.
\(a,2\frac{x}{7}=\frac{75}{35}\)
\(\frac{75}{35}=2\frac{5}{35}=2\frac{1}{7}\)
\(\Rightarrow x=1\)
\(b,47:4=...\left(+3\right)\)
\(\Rightarrow\left(47-3\right):4=11\)
\(\Rightarrow x=11\)
\(c,\frac{112}{5}=\frac{336}{15}=22\left(+6\right)\)
\(\Rightarrow x=22\frac{x=6}{15}\)
\(d,7,5x=\frac{75x}{100}:\left(9-6\frac{13}{21}\right)=2\frac{13}{25}\)
\(\frac{75x}{100}:\left(9-\frac{139}{21}\right)=\frac{63}{25}\)
\(\frac{75x}{100}:\frac{50}{21}=\frac{63}{25}\)
\(\Rightarrow\frac{63^3}{25^{\left(1\right)}}\cdot\frac{50^{\left(2\right)}}{21^1}=6=\frac{6}{1}\)
\(\Rightarrow\frac{6}{1}=\frac{600}{100}\)
Con d sai đề
Đè thừa một số \(\frac{25}{156}\),mk ko lại đề bài nhé
\(A=1-\frac{2+3}{2\cdot3}+.....+\frac{11+12}{11\cdot12}-\frac{12+13}{12\cdot13}\)
\(=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}-...+\frac{1}{11}+\frac{1}{12}-\frac{1}{12}-\frac{1}{13}\)
\(=\frac{1}{2}-\frac{1}{13}=\frac{11}{26}\)
a,\(=\frac{-5}{9}+\frac{8}{15}+\frac{-2}{11}+\frac{-4}{9}+\frac{7}{15}\)
\(\left(\frac{-5}{9}+\frac{-4}{9}\right)+\left(\frac{8}{15}+\frac{7}{15}\right)+\frac{-2}{11}\)
=-1+1+-2/11
=0+-2/11
=-2/11
b,\(=\left(\frac{5}{13}+\frac{8}{13}\right)+\left(\frac{-20}{41}+\frac{-21}{40}\right)+\frac{-5}{17}\)
=1+-1+-5/17
=0+-5/17
=-5/17
c,\(=\left(\frac{1}{5}+\frac{4}{5}\right)+\left(\frac{-2}{9}+-\frac{7}{9}\right)+\frac{16}{17}\)
=1+-1+16/17
=0+16/17
=16/17
d,\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)
a.\(\frac{-5}{9}\)+\(\frac{8}{15}\)+\(\frac{-2}{11}\)+\(\frac{4}{-9}\)+\(\frac{7}{15}\)
=\(\frac{-5}{9}\)+\(\frac{4}{-9}\)+\(\frac{8}{15}\)+\(\frac{7}{15}\)+\(\frac{-2}{11}\)
=(\(\frac{-5}{9}\)+\(\frac{-4}{9}\))+(\(\frac{8}{15}\)+\(\frac{7}{15}\))+\(\frac{-2}{11}\)
=(-1)+1+\(\frac{-2}{11}\)
=0+\(\frac{-2}{11}\)
=\(\frac{-2}{11}\).