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11/2 - 19/4
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\(\dfrac{11}{2}-\left(\dfrac{7}{5}-\dfrac{1}{2}\right)=\dfrac{11}{2}-\left(\dfrac{14}{10}-\dfrac{5}{10}\right)=\dfrac{11}{2}-\dfrac{9}{10}=\dfrac{55}{10}-\dfrac{9}{10}=\dfrac{46}{10}=\dfrac{23}{5}\)
\(a,11\dfrac{3}{4}-\left(6\dfrac{5}{6}-4\dfrac{1}{2}\right)+1\dfrac{2}{3}\)
\(=\dfrac{47}{4}-\left(\dfrac{41}{6}-\dfrac{9}{2}\right)+\dfrac{5}{3}\)
\(=\dfrac{47}{4}-\left(\dfrac{41}{6}-\dfrac{27}{6}\right)+\dfrac{5}{3}\)
\(=\dfrac{47}{4}-\dfrac{14}{6}+\dfrac{5}{3}\)
\(=\dfrac{47}{4}-\dfrac{7}{3}+\dfrac{5}{3}\)
\(=\dfrac{47}{4}-\left(\dfrac{7}{3}+\dfrac{5}{3}\right)\)
\(=\dfrac{47}{4}-\dfrac{12}{3}\)
\(=\dfrac{47}{4}-4\)
\(=\dfrac{47}{4}-\dfrac{16}{4}\)
\(=\dfrac{31}{4}\)
c) Ta có: \(4\dfrac{3}{7}:\left(\dfrac{7}{5}\cdot4\dfrac{3}{7}\right)\)
\(=\dfrac{31}{7}:\left(\dfrac{7}{5}\cdot\dfrac{31}{7}\right)\)
\(=\dfrac{31}{7}:\dfrac{31}{5}\)
\(=\dfrac{5}{7}\)
d)\(\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{32}{99}\)
a) 11/4x(-0,4)x-1,6x11/4
= -1,1 x-1,6x11/4
= 1,76x11/4
= 4,48
b) 6/19x-7/11+6/19x-4/19x-4/11+13/19
= -42/209+96/3971+13/19
= -702/3971+13/19
= 2015/3971
c) 2/5x3/7+-3/7x3/5+3/7
= 3/7x(2/5+3/5+1)
= 3/7x(1+1)
= 3/7x2
= 6/7
d) 2/3x5+2x5/7+2/7x9+....+2/97x99
= 1/3-1/5+1/5-1/7+1/7-1/9+......+1/97-1/99
= 1/3-1/99
= 32/99
a: \(\left(-\dfrac{5}{6}+\dfrac{2}{5}\right):\dfrac{3}{8}+\left(\dfrac{4}{5}-\dfrac{11}{30}\right):\dfrac{3}{8}\)
\(=\left(-\dfrac{5}{6}+\dfrac{2}{5}\right)\cdot\dfrac{8}{3}+\left(\dfrac{4}{5}-\dfrac{11}{30}\right)\cdot\dfrac{8}{3}\)
\(=\dfrac{8}{3}\left(-\dfrac{5}{6}+\dfrac{2}{5}+\dfrac{4}{5}-\dfrac{11}{30}\right)\)
\(=\dfrac{8}{3}\cdot\dfrac{-25+36-11}{30}\)
=0
b: \(\left(-\dfrac{3}{4}+\dfrac{2}{5}\right):\dfrac{3}{7}+\left(\dfrac{3}{5}+\dfrac{-1}{4}\right):\dfrac{3}{7}\)
\(=\left(-\dfrac{3}{4}+\dfrac{2}{5}\right)\cdot\dfrac{7}{3}+\left(\dfrac{3}{5}-\dfrac{1}{4}\right)\cdot\dfrac{7}{3}\)
\(=\dfrac{7}{3}\left(-\dfrac{3}{4}+\dfrac{2}{5}+\dfrac{3}{5}-\dfrac{1}{4}\right)\)
\(=\dfrac{7}{3}\cdot0=0\)
c: \(\dfrac{-13}{18}\cdot\dfrac{5}{8}+\dfrac{-5}{18}\cdot\dfrac{2}{9}+\dfrac{-13}{18}\cdot\dfrac{3}{8}+\dfrac{-5}{18}\cdot\dfrac{7}{9}\)
\(=\left(-\dfrac{13}{18}\cdot\dfrac{5}{8}+\dfrac{-13}{18}\cdot\dfrac{3}{8}\right)+\left(-\dfrac{5}{18}\cdot\dfrac{2}{9}+\dfrac{-5}{18}\cdot\dfrac{7}{9}\right)\)
\(=-\dfrac{13}{18}\left(\dfrac{5}{8}+\dfrac{3}{8}\right)+\dfrac{-5}{18}\left(\dfrac{2}{9}+\dfrac{7}{9}\right)\)
\(=-\dfrac{13}{18}-\dfrac{5}{18}=-\dfrac{18}{18}=-1\)
d: Sửa đề: \(\dfrac{-11}{19}\cdot\dfrac{4}{9}+\dfrac{-8}{19}\cdot\dfrac{3}{7}+\dfrac{-11}{19}\cdot\dfrac{5}{9}+\dfrac{-8}{19}\cdot\dfrac{4}{7}\)
\(=\left(-\dfrac{11}{19}\cdot\dfrac{4}{9}+\dfrac{-11}{19}\cdot\dfrac{5}{9}\right)+\left(\dfrac{-8}{19}\cdot\dfrac{3}{7}+\dfrac{-8}{19}\cdot\dfrac{4}{7}\right)\)
\(=-\dfrac{11}{19}\left(\dfrac{4}{9}+\dfrac{5}{9}\right)+\dfrac{-8}{19}\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\)
\(=-\dfrac{11}{19}-\dfrac{8}{19}=-\dfrac{19}{19}=-1\)
\(a.\left(-\dfrac{5}{6}+\dfrac{2}{5}\right):\dfrac{3}{8}+\left(\dfrac{4}{5}-\dfrac{11}{30}\right):\dfrac{3}{8}\)
\(=\left(-\dfrac{13}{30}\right):\dfrac{3}{8}+\dfrac{13}{30}:\dfrac{3}{8}\)
\(=\left[\left(-\dfrac{13}{30}+\dfrac{13}{30}\right)\right]:\dfrac{3}{8}\)
\(=0:\dfrac{3}{8}=0\)
\(b.\left(-\dfrac{3}{4}+\dfrac{2}{5}\right):\dfrac{3}{7}+\left(\dfrac{3}{5}+-\dfrac{1}{4}\right):\dfrac{3}{7}\)
\(=\left(-\dfrac{7}{20}\right):\dfrac{3}{7}+\dfrac{7}{20}:\dfrac{3}{7}\)
\(=\left[\left(-\dfrac{7}{20}+\dfrac{7}{20}\right)\right]:\dfrac{3}{7}=0:\dfrac{3}{7}=0\)
\(c.-\dfrac{13}{18}.\dfrac{5}{8}+-\dfrac{5}{18}.\dfrac{2}{9}+-\dfrac{13}{18}.\dfrac{3}{8}+-\dfrac{5}{18}.\dfrac{7}{9}\)
\(=\left(\dfrac{5}{8}+\dfrac{3}{8}\right).-\dfrac{13}{18}+\left(\dfrac{2}{9}+\dfrac{7}{9}\right).-\dfrac{5}{18}\)
\(=1.-\dfrac{13}{18}+1.-\dfrac{5}{18}=-\dfrac{13}{18}+-\dfrac{5}{18}=-1\)
\(d.-\dfrac{11}{19}.\dfrac{4}{9}+\dfrac{-8}{19}.\dfrac{3}{7}+-\dfrac{11}{19}.\dfrac{5}{9}+-\dfrac{9}{19}.\dfrac{4}{7}\)
\(=\left(\dfrac{4}{9}+\dfrac{5}{9}\right).-\dfrac{11}{19}+-\dfrac{24}{133}+-\dfrac{36}{133}\)
\(=-\dfrac{11}{19}+-\dfrac{60}{133}=-\dfrac{137}{133}\)
em ơi đây là toán nâng cao lớp 5 nhé lớp em chưa làm được đâu nhé
1) Ta có: \(\frac{-4}{7}-\frac{11}{19}+\frac{13}{19}\cdot\frac{-3}{7}+\frac{2}{19}:\frac{-7}{4}\)
\(=\frac{-4}{7}-\frac{11}{19}-\frac{39}{133}-\frac{8}{133}\)
\(=\frac{-76}{133}-\frac{77}{133}-\frac{39}{133}-\frac{8}{133}\)
\(=\frac{-200}{133}\)
2) Ta có: \(\left(\frac{-4}{9}+\frac{3}{5}\right):\frac{1}{\frac{1}{5}}+\left(\frac{1}{5}-\frac{5}{9}\right):\frac{1}{\frac{1}{5}}\)
\(=\left(\frac{-4}{9}+\frac{3}{5}\right)\cdot\frac{1}{5}+\left(\frac{1}{5}-\frac{5}{9}\right)\cdot\frac{1}{5}\)
\(=\frac{1}{5}\left(\frac{-4}{9}+\frac{3}{5}+\frac{1}{5}-\frac{5}{9}\right)\)
\(=\frac{1}{5}\left(-1+\frac{4}{5}\right)\)
\(=\frac{1}{5}\cdot\frac{-1}{5}=\frac{-1}{25}\)
3) Ta có: \(\frac{4}{5}-\left(-\frac{2}{7}\right)-\frac{7}{10}\)
\(=\frac{4}{5}+\frac{2}{7}-\frac{7}{10}\)
\(=\frac{56}{70}+\frac{20}{70}-\frac{49}{70}\)
\(=\frac{27}{70}\)
4) Ta có: \(\frac{2}{7}-\left(-\frac{13}{15}+\frac{4}{9}\right)-\left(\frac{5}{9}-\frac{2}{15}\right)\)
\(=\frac{2}{7}+\frac{13}{15}-\frac{4}{9}-\frac{5}{9}+\frac{2}{15}\)
\(=\frac{2}{7}+1-1=\frac{2}{7}\)
a) 12 + 3 = 15 15 + 4 = 19 8 + 2 = 10 14 + 3 = 17
15 - 3 = 12 19 - 4 = 15 10 - 2 = 8 17 - 3 = 14
b) 11 + 4 + 2 = 17 19 - 5 - 4 = 10 14 + 2 - 5 = 11
a) Ta có:
\(A=\dfrac{-68}{123}\cdot\dfrac{-23}{79}=\dfrac{68}{123}\cdot\dfrac{23}{79}\)
\(B=\dfrac{-14}{79}\cdot\dfrac{-68}{7}\cdot\dfrac{-46}{123}=-\left(\dfrac{14}{79}\cdot\dfrac{68}{7}\cdot\dfrac{46}{123}\right)\)
\(C=\dfrac{-4}{19}\cdot\dfrac{-3}{19}\cdot...\cdot\dfrac{0}{19}\cdot...\cdot\dfrac{3}{19}\cdot\dfrac{4}{19}=0\)
Suy ra A là số hữu tỉ dương, B là số hữu tỉ âm và C là 0.
Vậy A > C > B.
b) Ta có:
\(\dfrac{B}{A}=\dfrac{-\left(\dfrac{14}{79}\cdot\dfrac{68}{7}\cdot\dfrac{46}{123}\right)}{\dfrac{68}{123}\cdot\dfrac{23}{79}}=-\dfrac{14}{79}\cdot\dfrac{68}{7}\cdot\dfrac{46}{123}\cdot\dfrac{123}{68}\cdot\dfrac{79}{23}\)
\(\dfrac{B}{A}=-\dfrac{14\cdot68\cdot46\cdot123\cdot79}{79\cdot7\cdot123\cdot68\cdot23}=-\left(2\cdot2\right)=-4\)
Vậy B : A = -4
\(\frac{11}{2}\)- \(\frac{19}{4}\)= \(\frac{22}{4}\)- \(\frac{19}{4}\)= \(\frac{3}{4}\)
\(\frac{11}{2}-\frac{19}{4}=\frac{3}{4}\)