\(\dfrac{3}{5}\cdot\dfrac{2}{7}+\dfrac{3}{5}\cdot\dfrac{9}{7}-\dfrac{3}{5}\cdot\dfrac{3}{7}\)
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\(e.\dfrac{7}{10}\cdot\dfrac{-3}{5}+\dfrac{7}{10}\cdot\dfrac{-2}{5}-\dfrac{3}{10}\)
\(=\dfrac{7}{10}\cdot\left[\left(\dfrac{-3}{5}\right)+\left(\dfrac{-2}{5}\right)\right]-\dfrac{3}{10}\)
\(=\dfrac{7}{10}\cdot1-\dfrac{3}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)
\(f.\dfrac{-3}{7}\cdot\dfrac{5}{9}+\dfrac{4}{9}\cdot\dfrac{-3}{7}+2\dfrac{3}{7}\)
\(=\dfrac{-3}{7}\cdot\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{17}{3}\)
\(=\dfrac{-3}{7}\cdot1+\dfrac{17}{3}=\dfrac{-9}{21}+\dfrac{119}{21}=\dfrac{110}{21}\)
\(g.\dfrac{5}{9}\cdot\dfrac{10}{17}+\dfrac{5}{9}\cdot\dfrac{9}{17}-\dfrac{5}{9}\cdot\dfrac{2}{17}\)
\(=\dfrac{5}{9}\cdot\left(\dfrac{10}{17}+\dfrac{9}{17}-\dfrac{2}{17}\right)\)
\(=\dfrac{5}{9}\cdot1=\dfrac{5}{9}\)

a) A=[27(14−13)]:[27(13−25)]=(14−13):(13−25)=114A=[27(14−13)]:[27(13−25)]=(14−13):(13−25)=114.
b) B=34(15−27−13+27)15(27+13)−13(27+13)=34(15−13)(15−13)(27+13)=11152B=34(15−27−13+27)15(27+13)−13(27+13)=34(15−13)(15−13)(27+13)=11152.
a) \mathrm{A}=\left[\dfrac{2}{7}\left(\dfrac{1}{4}-\dfrac{1}{3}\right)\right]:\left[\dfrac{2}{7}\left(\dfrac{1}{3}-\dfrac{2}{5}\right)\right]=\left(\dfrac{1}{4}-\dfrac{1}{3}\right):\left(\dfrac{1}{3}-\dfrac{2}{5}\right)=1 \dfrac{1}{4}A=[72(41−31)]:[72(31−52)]=(41−31):(31−52)=141.
b) \mathrm{B}=\dfrac{\dfrac{3}{4}\left(\dfrac{1}{5}-\dfrac{2}{7}-\dfrac{1}{3}+\dfrac{2}{7}\right)}{\dfrac{1}{5}\left(\dfrac{2}{7}+\dfrac{1}{3}\right)-\dfrac{1}{3}\left(\dfrac{2}{7}+\dfrac{1}{3}\right)}=\dfrac{\dfrac{3}{4}\left(\dfrac{1}{5}-\dfrac{1}{3}\right)}{\left(\dfrac{1}{5}-\dfrac{1}{3}\right)\left(\dfrac{2}{7}+\dfrac{1}{3}\right)}=1 \dfrac{11}{52}B=51(72+31)−31(72+31)43(51−72−31+72)=(51−31)(72+31)43(51−31)=15211

Bài 1:
\(a,=\frac{2}{3}-\frac{16}{3}=\frac{-14}{3}\)
\(b,=\left(\frac{3}{7}+\frac{4}{7}\right)+\left(-\frac{6}{19}+\frac{-13}{19}\right)=1-1=0\)
\(c,=\frac{3}{5}.\left(\frac{8}{9}-\frac{7}{9}+\frac{26}{9}\right)=\frac{3}{5}.3=\frac{9}{5}\)
a,\(\dfrac{1}{2}\).\(\dfrac{4}{3}\)-\(\dfrac{20}{3}\).\(\dfrac{4}{5}\)=\(\dfrac{2}{3}\)-\(\dfrac{16}{3}\)=-\(\dfrac{14}{3}\)

Giải:
a) \(\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{-\left(\dfrac{4}{5}+\dfrac{1}{3}\right).\dfrac{1}{2}+1}=2\dfrac{33}{52}\)
\(\Leftrightarrow\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{-\dfrac{17}{15}.\dfrac{1}{2}+1}=\dfrac{137}{52}\)
\(\Leftrightarrow\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{\dfrac{13}{30}}=\dfrac{137}{52}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}=\dfrac{137}{52}.\dfrac{13}{30}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}=\dfrac{137}{120}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}=\dfrac{137}{120}+\dfrac{1}{6}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}=\dfrac{157}{120}\)
\(\Leftrightarrow x+\dfrac{3}{4}=\dfrac{157}{120}:\dfrac{7}{2}\)
\(\Leftrightarrow x+\dfrac{3}{4}=\dfrac{157}{420}\)
\(\Leftrightarrow x=\dfrac{157}{420}-\dfrac{3}{4}\)
\(\Leftrightarrow x=-\dfrac{79}{210}\)
Vậy \(x=-\dfrac{79}{210}\).
b) \(\dfrac{\left(5-\dfrac{2}{7}\right).\dfrac{7}{9}.\dfrac{3}{5}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=5\dfrac{5}{21}\)
\(\Leftrightarrow\dfrac{\left(5-\dfrac{2}{7}\right).\dfrac{7}{15}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)
\(\Leftrightarrow\dfrac{\dfrac{33}{7}.\dfrac{7}{15}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)
\(\Leftrightarrow\dfrac{\dfrac{11}{5}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)
\(\Leftrightarrow\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}=\dfrac{11}{5}:\dfrac{110}{21}\)
\(\Leftrightarrow\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}=\dfrac{21}{50}\)
\(\Leftrightarrow3x-\dfrac{5}{6}=\dfrac{21}{50}.\dfrac{1}{7}\)
\(\Leftrightarrow3x-\dfrac{5}{6}=\dfrac{3}{50}\)
\(\Leftrightarrow3x=\dfrac{3}{50}+\dfrac{5}{6}\)
\(\Leftrightarrow3x=\dfrac{67}{75}\)
\(\Leftrightarrow x=\dfrac{67}{75}:3\)
\(\Leftrightarrow x=\dfrac{67}{225}\)
Vậy \(x=\dfrac{67}{225}\).
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a, \(\begin{array}{l}A = \dfrac{{ - 3}}{{14}} + \dfrac{2}{{13}} + \dfrac{{ - 25}}{{14}} + \dfrac{{ - 15}}{{13}}\\A = \left( {\dfrac{{ - 3}}{{14}} + \dfrac{{ - 25}}{{14}}} \right) + \left( {\dfrac{2}{{13}} + \dfrac{{ - 15}}{{13}}} \right)\\A = \dfrac{{ - 3 + \left( { - 25} \right)}}{{14}} + \dfrac{{2 + \left( { - 15} \right)}}{{13}}\\A = \dfrac{{ - 28}}{{14}} + \dfrac{{ - 13}}{{13}}\\A = - 2 + (-1)\\A = - 3\end{array}\)
b,
Cách 1:
\(\begin{array}{l}B = \dfrac{5}{3}.\dfrac{7}{{25}} + \dfrac{5}{3}.\dfrac{{21}}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}\\B = \left( {\dfrac{5}{3}.\dfrac{7}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}} \right) + \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = 0 + \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = \dfrac{{5.21}}{{3.25}}\\B = \dfrac{7}{5}\end{array}\)
Cách 2:
\(B = \dfrac{5}{3}.\dfrac{7}{{25}} + \dfrac{5}{3}.\dfrac{{21}}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}\\B = \dfrac{5}{3}.({\dfrac{7}{{25}} -\dfrac{7}{{25}} + \dfrac{{21}}{{25}}})\\B = \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = \dfrac{{5.21}}{{3.25}}\\B = \dfrac{7}{5}\)

a: =3/4-5/4+3/2x12/5=-1/2+36/10=-5/10+36/10=31/10
b: \(=\dfrac{-7}{13}\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{7}{13}+8=8\)

\(\sqrt{\dfrac{16}{49}}+\left(\dfrac{1}{2}\right)^3-\left|-\dfrac{4}{7}\right|-\dfrac{7}{8}\)
\(=\dfrac{4}{7}+\dfrac{1}{8}-\dfrac{4}{7}-\dfrac{7}{8}\)
\(=\dfrac{1}{8}-\dfrac{7}{8}=-\dfrac{6}{8}=-\dfrac{3}{4}\)
\(\left|\dfrac{1}{2}-\dfrac{3}{5}\right|\cdot\sqrt{9}+0,5\left(-2\dfrac{3}{5}\right)\)
\(=\left|\dfrac{5-6}{10}\right|\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)
\(=\dfrac{1}{10}\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)
\(=\dfrac{3}{10}-\dfrac{13}{10}=-\dfrac{10}{10}=-1\)

Hoàng Ngọc Anh bài này nè bn giúp mk nha!!! ngày mai mk phải nộp bài rùi =.=
a) \(\Rightarrow\dfrac{\dfrac{7}{2}x+\dfrac{59}{24}}{\dfrac{13}{30}}=\dfrac{137}{52}\)
\(\Rightarrow\left(\dfrac{7}{2}x+\dfrac{59}{24}\right).52=\dfrac{13}{30}.137\)
\(\Rightarrow182x+\dfrac{767}{6}=\dfrac{1781}{30}\)
\(\Rightarrow x=\dfrac{-79}{210}\)
b) Tương tự câu a)

a) A=35.67+37.35−27.35A=35.67+37.35−27.35
=35⋅(67+37−27)=35=35⋅(67+37−27)=35
b) B=(−13⋅25+−29⋅25+25⋅119)⋅52B=(−13⋅25+−29⋅25+25⋅119)⋅52
=(−13−29+119)⋅25⋅52=−13+(119−29)=−12.=(−13−29+119)⋅25⋅52=−13+(119−29)=−12.
c) C=(−45+57)⋅32+(−15+27)⋅32=(−45+57+−15+27)⋅32=((−45+−15)+(57+27))⋅32=0.C=(−45+57)⋅32+(−15+27)⋅32=(−45+57+−15+27)⋅32=((−45+−15)+(57+27))⋅32=0.
d) D=49:(115−1015)+49:(222−522)D=49:(115−1015)+49:(222−522)
=49:−35+49:−322=49⋅−53+49.−223=49:−35+49:−322=49⋅−53+49.−223
=49⋅(−53+−223)=49.−273=−4.
a) \mathrm{A}=\dfrac{3}{5}. \dfrac{6}{7}+\dfrac{3}{7}. \dfrac{3}{5}-\dfrac{2}{7}. \dfrac{3}{5}A=53. 76+73. 53−72. 53
b) \mathrm{B} =\left(-13 \cdot \dfrac{2}{5}+\dfrac{-2}{9} \cdot \dfrac{2}{5}+\dfrac{2}{5} \cdot \dfrac{11}{9}\right) \cdot \dfrac{5}{2} B=(−13⋅52+9−2⋅52+52⋅911)⋅25
=\left(-13-\dfrac{2}{9}+\dfrac{11}{9}\right) \cdot \dfrac{2}{5} \cdot \dfrac{5}{2}=-13+\left(\dfrac{11}{9}-\dfrac{2}{9}\right)=-12 .=(−13−92+911)⋅52⋅25=−13+(911−92)=−12.
c) \mathrm{C} =\left(\dfrac{-4}{5}+\dfrac{5}{7}\right) \cdot \dfrac{3}{2}+\left(\dfrac{-1}{5}+\dfrac{2}{7}\right) \cdot \dfrac{3}{2} =\left(\dfrac{-4}{5}+\dfrac{5}{7}+\dfrac{-1}{5}+\dfrac{2}{7}\right) \cdot \dfrac{3}{2}=\left(\left(\dfrac{-4}{5}+\dfrac{-1}{5}\right)+\left(\dfrac{5}{7}+\dfrac{2}{7}\right)\right) \cdot \dfrac{3}{2}=0 .C=(5−4+75)⋅23+(5−1+72)⋅23=(5−4+75+5−1+72)⋅23=((5−4+5−1)+(75+72))⋅23=0.
d) \mathrm{D}=\dfrac{4}{9}:\left(\dfrac{1}{15}-\dfrac{10}{15}\right)+\dfrac{4}{9}:\left(\dfrac{2}{22}-\dfrac{5}{22}\right)D=94:(151−1510)+94:(222−225)
\(\dfrac{3}{5}\cdot\dfrac{2}{7}+\dfrac{3}{5}\cdot\dfrac{9}{7}-\dfrac{3}{5}\cdot\dfrac{3}{7}=\dfrac{3}{5}\cdot\left(\dfrac{2}{7}+\dfrac{9}{7}-\dfrac{3}{7}\right)=\dfrac{3}{5}\cdot\dfrac{8}{7}=\dfrac{24}{35}\)