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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=\(\frac{-1}{2}\)
a: \(=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}+\dfrac{-\dfrac{1}{4}\cdot\dfrac{-2}{3}-\dfrac{3}{4}:\dfrac{1}{6}}{\dfrac{3}{2}\cdot\left(\dfrac{-2}{3}-\dfrac{3}{4}\cdot\dfrac{-2}{3}\right)}\)
\(=\dfrac{3}{4}+\dfrac{\dfrac{2}{12}-\dfrac{9}{2}}{\dfrac{3}{2}\cdot\dfrac{-1}{6}}=\dfrac{3}{4}+\dfrac{-13}{3}:\dfrac{-3}{12}\)
\(=\dfrac{3}{4}+\dfrac{13}{3}\cdot\dfrac{12}{3}=\dfrac{3}{4}+\dfrac{156}{9}=\dfrac{217}{12}\)
b: \(A=158\left(\dfrac{12\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}{4\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}:\dfrac{5\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}{6\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}\right)\cdot\dfrac{50550505}{711711711}\)
\(=158\cdot\left(3\cdot\dfrac{6}{5}\right)\cdot\dfrac{50550505}{711711711}\)
\(\simeq40.39\)
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.
Đè thừa một số \(\frac{25}{156}\),mk ko lại đề bài nhé
\(A=1-\frac{2+3}{2\cdot3}+.....+\frac{11+12}{11\cdot12}-\frac{12+13}{12\cdot13}\)
\(=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}-...+\frac{1}{11}+\frac{1}{12}-\frac{1}{12}-\frac{1}{13}\)
\(=\frac{1}{2}-\frac{1}{13}=\frac{11}{26}\)