a)\(\dfrac{-1}{3}+\dfrac{2}{1}-\dfrac{6}{5}=\dfrac{-5}{15}+\dfrac{30}{15}-\dfrac{18}{15}=\dfrac{7}{15}\)

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a, 3 \(\frac{14}{19}\)+ \(\frac{13}{17}\)+ \(\frac{35}{43}\)+ 6\(\frac{5}{19}\)+ \(\frac{8}{43}\)= \(\left(3\frac{14}{19}+6\frac{5}{19}\right)+\left(\frac{35}{43}+\frac{8}{43}\right)+\frac{13}{17}=\)\(9+1+\frac{13}{17}=8+\frac{13}{17}=8\frac{13}{17}\)

b, \(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+1\frac{5}{7}\)\(=\frac{-5}{7}\left(\frac{2}{11}+\frac{9}{11}\right)+1\frac{5}{7}\)\(=\frac{-5}{7}.1+1\frac{5}{7}\)\(=\frac{-5}{7}+\frac{12}{7}=\frac{7}{7}=1\)

Chúc bn học tốt

\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\frac{5}{19}+\frac{8}{43}\)

\(=\left(3\frac{14}{19}+6\frac{5}{19}\right)+\left(\frac{35}{43}+\frac{8}{43}\right)+\frac{13}{17}\)

\(=10+1+\frac{13}{17}=11+\frac{13}{17}=11\frac{13}{17}\)

a)

\(3\dfrac{14}{19}+\dfrac{13}{17}+\dfrac{35}{43}+6\dfrac{5}{19}+\dfrac{8}{43}\\ =\left(3\dfrac{14}{19}+6\dfrac{5}{19}\right)+\left(\dfrac{35}{43}+\dfrac{8}{43}\right)+\dfrac{13}{17}\\ =10+1+\dfrac{13}{17}\\ =11\dfrac{13}{17}\)

b)

\(\dfrac{-5}{7}\cdot\dfrac{2}{11}+\dfrac{-5}{7}\cdot\dfrac{9}{11}+1\dfrac{5}{7}\\ =\dfrac{-5}{7}\cdot\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1\dfrac{5}{7}\\ =\dfrac{-5}{7}\cdot1+1\dfrac{5}{7}\\ =\dfrac{-5}{7}+1\dfrac{5}{7}\\ =1\)

a) \(3\dfrac{14}{19}+\dfrac{13}{17}+\dfrac{35}{43}+6\dfrac{5}{19}+\dfrac{8}{43}\)

\(=\left(3\dfrac{14}{19}+6\dfrac{5}{19}\right)+\left(\dfrac{35}{43}+\dfrac{8}{43}\right)+\dfrac{13}{17}\)

\(=\left[\left(3+6\right)+\left(\dfrac{14}{19}+\dfrac{5}{19}\right)\right]+1+\dfrac{13}{17}\)

\(=\left[9+1\right]+1+\dfrac{13}{17}\)

\(=10+1+\dfrac{13}{17}\)

\(=11+\dfrac{13}{17}\)

\(=\dfrac{187}{17}+\dfrac{13}{17}\)

\(=\dfrac{200}{17}\)

b) \(\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\)

\(=\dfrac{-5}{7}.\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\)

\(=\dfrac{-5}{7}.1+\dfrac{12}{7}\)

\(=\dfrac{-5}{7}+\dfrac{12}{7}\)

\(=\dfrac{7}{7}\)

\(=1\)

c) \(11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)

= \(11\dfrac{3}{13}-2\dfrac{4}{7}-5\dfrac{3}{13}\)

\(=\left(11\dfrac{3}{13}-5\dfrac{3}{13}\right)-2\dfrac{4}{7}\)

\(=\left[\left(11-5\right)+\left(\dfrac{3}{13}-\dfrac{3}{13}\right)\right]-\dfrac{18}{7}\)

\(=\left[6+0\right]-\dfrac{18}{7}\)

\(=6-\dfrac{18}{7}\)

\(=\dfrac{42}{7}-\dfrac{18}{7}\)

\(=\dfrac{24}{7}\)

d) \(\dfrac{2}{7}.5\dfrac{1}{4}-\dfrac{2}{7}.3\dfrac{1}{4}\)

\(=\dfrac{2}{7}.\left(5\dfrac{1}{4}-3\dfrac{1}{4}\right)\)

\(=\dfrac{2}{7}.\left[\left(5-3\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)\right]\)

\(=\dfrac{2}{7}.\left[2+0\right]\)

\(=\dfrac{2}{7}.2\)

= \(\dfrac{4}{7}\)

\(\dfrac{\dfrac{1}{4}\left(4,75-\dfrac{3}{20}\right).3,2}{0,23:\dfrac{5}{8}.0,5}+\left|-0,75\right|.2\dfrac{1}{3}\)

\(=\left[\dfrac{1}{4}\left(4,75-\dfrac{3}{20}\right).3,2\right]:\left[0,23:\dfrac{5}{8}.0,5\right]+0,75.2\dfrac{1}{3}\)

\(=\left(\dfrac{1}{4}.\dfrac{23}{5}.\dfrac{16}{5}\right):\left(\dfrac{23}{100}:\dfrac{5}{8}.\dfrac{1}{2}\right)+\dfrac{3}{8}\)

\(=\dfrac{92}{25}:\dfrac{23}{125}+\dfrac{3}{8}=20+\dfrac{3}{8}=\dfrac{163}{8}\)

\(A=\frac{5}{7}\cdot\frac{2}{3}\cdot\frac{7}{5}\cdot\frac{3}{2}\cdot\frac{19}{43}\)

   \(=(\frac{5}{7}\cdot\frac{7}{5})\cdot(\frac{2}{3}\cdot\frac{3}{2})\cdot\frac{19}{43}\)

    \(=1\cdot1\cdot\frac{19}{43}=\frac{19}{43}\)

A=5/7.2/3.7/5.3/2.19/43

A=5.2.7.3.19/7.3.5.2.43

A=19/43

Nhớ k cho mình nhé

Giúp Mik ik mai nộp oy

\(\frac27\times5\frac14-\frac27\times3\frac14\)

=\(\frac27\times\left(5\frac14-3\frac14\right)\)

=\(\frac27\times\left(\left(5-3\right)+\left(\frac14-\frac15\right)\right)\)

=\(\frac27\times\left(2+0\right)\)

=\(\frac27\times2\)

=\(\frac47\)

\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\frac{5}{19}+\frac{8}{43}\)

\(=\left(3\frac{14}{19}+6\frac{5}{19}\right)+\left(\frac{35}{43}+\frac{8}{43}\right)+\frac{13}{17}\)

\(=10+1+\frac{13}{17}=11+\frac{13}{17}=11\frac{13}{17}\)

\(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+1\frac{5}{7}\)

\(=\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+1+\frac{-5}{7}.\left(-1\right)\)

\(=\frac{-5}{7}\left(\frac{2}{11}+\frac{9}{11}-1\right)+1\)

\(=\frac{-5}{7}.0+1==0+1=1\)

Sửa đề là chứng minh nha bạn.

Ta có: \(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{80}>\dfrac{1}{41}+\dfrac{1}{41}+\dfrac{1}{41}+...+\dfrac{1}{41}\)(40 phân số \(\dfrac{1}{41}\))

\(=\dfrac{1.40}{41}=\dfrac{40}{41}>\dfrac{7}{12}\) (*)

Từ (*) suy ra: \(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{80}>\dfrac{7}{12}^{\left(đpcm\right)}\)

đpcm là gì