a: =27/45-20/45=7/45

b: \(=\dfrac{3}{5}+\dfrac{30}{40}=\dfrac{3}{5}+\dfrac{3}{4}=\dfrac{12}{20}+\dfrac{15}{20}=\dfrac{27}{20}\)

c: \(=\dfrac{8}{13}\left(\dfrac{7}{2}-\dfrac{5}{2}+1\right)=\dfrac{8}{13}\cdot2=\dfrac{16}{13}\)

d: \(=\dfrac{9}{23}\left(\dfrac{5}{17}-\dfrac{22}{17}\right)+11+\dfrac{9}{23}=11\)

a) \(\dfrac{3}{5}+\dfrac{-4}{9}=\dfrac{27}{45}+\dfrac{-20}{45}=\dfrac{7}{45}\)

b) \(\dfrac{3}{5}+\dfrac{2}{5}.\dfrac{15}{8}=1.\dfrac{15}{8}=\dfrac{15}{8}\)

c) \(\dfrac{7}{2}.\dfrac{8}{13}+\dfrac{8}{13}.\dfrac{-5}{2}+\dfrac{8}{13}=\dfrac{8}{13}.\left(\dfrac{7}{2}+\dfrac{-5}{2}\right)=\dfrac{8}{13}.1=\dfrac{8}{13}\)

d) \(\dfrac{-5}{17}.\dfrac{-9}{23}+\dfrac{9}{23}.\dfrac{-22}{17}+11\dfrac{9}{23}=\dfrac{9}{23}.\left(\dfrac{-5}{17}+\dfrac{-22}{17}\right)=\dfrac{-243}{391}\)

a) \(\dfrac{6}{{13}}.\dfrac{8}{{7}}.\dfrac{{ - 26}}{3}.\dfrac{{ - 7}}{8}\)

\(\begin{array}{l} = \left( {\dfrac{6}{{13}}.\dfrac{{ - 26}}{3}} \right).\left( {\dfrac{8}{7}.\dfrac{{ - 7}}{8}} \right)\\ = \dfrac{{6.\left( { - 26} \right)}}{{13.3}}.\dfrac{{8.\left( { - 7} \right)}}{{7.8}}\\=  (- 4).\left( { - 1} \right) = 4\end{array}\)

b) \(\dfrac{6}{5}.\dfrac{3}{{13}} - \dfrac{6}{5}.\dfrac{{16}}{{13}}\)

\(\begin{array}{l} = \dfrac{6}{5}.\left( {\dfrac{3}{{13}} - \dfrac{{16}}{{13}}} \right)\\ = \dfrac{6}{5}.\dfrac{{3 - 16}}{{13}}\\ = \dfrac{6}{5}.\dfrac{{-13}}{{13}}\\= \dfrac{6}{5}.\left( { - 1} \right)\\ = \dfrac{{ - 6}}{5}\end{array}\) 

a) Ta có: \(\dfrac{2}{3}x-1=\dfrac{3}{2}\)

\(\Leftrightarrow x\cdot\dfrac{2}{3}=\dfrac{5}{2}\)

hay \(x=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{5}{2}\cdot\dfrac{3}{2}=\dfrac{15}{4}\)

b) Ta có: \(\left|5x-\dfrac{1}{2}\right|-\dfrac{2}{7}=25\%\)

\(\Leftrightarrow\left|5x-\dfrac{1}{2}\right|=\dfrac{1}{4}+\dfrac{2}{7}=\dfrac{15}{28}\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{1}{2}=\dfrac{15}{28}\\5x-\dfrac{1}{2}=\dfrac{-15}{28}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{29}{28}\\5x=\dfrac{-1}{28}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{140}\\x=\dfrac{-1}{140}\end{matrix}\right.\)

c) Ta có: \(\dfrac{x-3}{4}=\dfrac{16}{x-3}\)

\(\Leftrightarrow\left(x-3\right)^2=64\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=8\\x-3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\)

d) Ta có: \(\dfrac{-8}{13}+\dfrac{7}{17}+\dfrac{21}{31}\le x\le\dfrac{-9}{14}+4-\dfrac{5}{14}\)

\(\Leftrightarrow\dfrac{3246}{6851}\le x\le3\)

\(\Leftrightarrow x\in\left\{1;2;3\right\}\)

ko biết =))

\(A=\left(-\dfrac{43}{51}\right)\left(-\dfrac{19}{80}\right)\)

=>A>0(1)

\(B=\left(-\dfrac{7}{13}\right)\left(-\dfrac{4}{65}\right)\left(-\dfrac{8}{21}\right)\)

=>B<0(2)

C\(=-\dfrac{5}{10}.\left(-\dfrac{4}{10}\right).....\left(\dfrac{4}{10}\right)\left(\dfrac{5}{10}\right)=0\)

=>C=0(3)

Từ 1;2;3 =>A>C>B

\(A=\dfrac{-43}{51}.\dfrac{-19}{80}\Leftrightarrow A>0\left(1\right)\)

\(B=\left(\dfrac{-7}{13}\right).\left(-\dfrac{4}{65}\right).\left(\dfrac{-8}{31}\right)\Leftrightarrow B< 0\left(2\right)\)

\(C=\dfrac{-5}{10}.\dfrac{-4}{10}...........\dfrac{3}{10}.\dfrac{4}{10}.\dfrac{5}{10}\Leftrightarrow C=0\left(3\right)\)

Từ \(\left(1\right)+\left(2\right)+\left(3\right)\Leftrightarrow A>C>B\)

\(a,\left(31\dfrac{6}{13}+5\dfrac{9}{41}\right)-36\dfrac{6}{13}\\ =\left(31-36\right)+\left(\dfrac{6}{13}-\dfrac{6}{13}\right)+5\dfrac{9}{41}\\ =-5+0+5\dfrac{9}{41}\\ =\left(-5+5\right)+\dfrac{9}{41}=\dfrac{9}{41}\)

\(b,\dfrac{5}{3}+\left(-\dfrac{2}{7}\right)-\left(-1,2\right)\\ =\dfrac{5}{3}-\dfrac{2}{7}+\dfrac{6}{5}\\ =\dfrac{5.35-2.15+6.21}{105}=\dfrac{271}{105}\\ c,0,25+\dfrac{3}{5}-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)=\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{1}{8}+\dfrac{2}{5}-1\dfrac{1}{4}\\ =\left(-1\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(\dfrac{3}{5}+\dfrac{2}{5}\right)-\dfrac{1}{8}=-1+1-\dfrac{1}{8}=-\dfrac{1}{8}\)

a) (31 6/13 + 5 9/41) - 36 6/13

= 409/13 + 214/41 - 474/13

= (409/13 - 474/13) + 214/41

= -5 + 214/41

= 9/41

b) 5/3 + (-2/7) - (-1,2)

= 5/3 - 2/7 + 6/5

= 29/21 + 6/5

= 271/105

c) 0,25 + 3/5 - (1/8 - 2/5 + 1 1/4)

= 1/4 + 3/5 - 1/8 + 2/5 - 5/4

= (1/4 - 5/4) + (3/5 + 2/5) - 1/8

= -1 + 1 - 1/8

= -1/8

\(\dfrac{2}{3}+\dfrac{1}{5}.\dfrac{10}{7}=\dfrac{2}{3}+\dfrac{10}{35}=\dfrac{70}{105}+\dfrac{30}{105}=\dfrac{100}{105}=\dfrac{50}{21}\)

a) Ta có: \(\dfrac{2}{3}+\dfrac{1}{5}\cdot\dfrac{10}{7}\)

\(=\dfrac{2}{3}+\dfrac{2}{7}\)

\(=\dfrac{14}{21}+\dfrac{6}{21}\)

\(=\dfrac{20}{21}\)

\(\dfrac{7}{12}-\dfrac{27}{7}.\dfrac{1}{8}=\dfrac{7}{12}-\dfrac{27}{56}=\dfrac{392}{672}-\dfrac{324}{672}=\dfrac{68}{672}=\dfrac{17}{168}\)

\(\dfrac{5}{2x1}+\dfrac{4}{1x11}+\dfrac{3}{11x2}+\dfrac{1}{2x15}+\dfrac{13}{15x4}+\dfrac{15}{4x13}\)

=7x(\(\dfrac{5}{2x7}+\dfrac{4}{7x11}+\dfrac{3}{11x14}+\dfrac{1}{14x15}+\dfrac{13}{15x28}+\dfrac{15}{28x43}\))

=7x\(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{28}+\dfrac{1}{28}-\dfrac{1}{43}\)=7x(\(\dfrac{1}{2}-\dfrac{1}{43}\))

=7x\(\dfrac{41}{86}\)

=\(\dfrac{287}{86}\)

5/2x1+4/1x11+3/11x2+1/2x15+13/15x4+15/4x43=7x(5/2x7+4/7x11+3/11x14+1/14x15+13/15x28+15/28x43)=7x(1/2-1/7+1/7-1/11+1/11-1/14+1/14+1/15+1/15-1/28+1/28-1/43)=7x(1/2-1/43)=7x41/86=287/86

a: \(\dfrac{7}{2}+5=\dfrac{7}{2}+\dfrac{10}{2}=\dfrac{17}{2}\)

b: \(9+\dfrac{12}{13}=\dfrac{117}{13}+\dfrac{12}{13}=\dfrac{129}{13}\)

c: \(8-\dfrac{9}{11}=\dfrac{88}{11}-\dfrac{9}{11}=\dfrac{88-9}{11}=\dfrac{79}{11}\)

d: \(\dfrac{21}{5}-3=\dfrac{21}{5}-\dfrac{15}{5}=\dfrac{6}{5}\)