giúp mk với

mai mk đi học rùi

a) _{^{\(\dfrac{3}{5}+\dfrac{11}{20}=\dfrac{12}{20}+\dfrac{11}{20}=\dfrac{23}{20}\)}}

b) _{^{\(\dfrac{5}{8}-\dfrac{4}{9}=\dfrac{45}{72}-\dfrac{32}{72}=\dfrac{13}{72}\)}}

c) _{^{\(\dfrac{9}{16}\times\dfrac{4}{3}=\dfrac{3}{4}\)}}

d) _{^{\(\dfrac{4}{7}:\dfrac{8}{11}=\dfrac{4}{7}\times\dfrac{11}{8}=\dfrac{11}{14}\)}}

e) _{^{\(\dfrac{3}{5}+\dfrac{4}{5}:\dfrac{2}{5}=\dfrac{3}{5}+\dfrac{4}{5}\times\dfrac{5}{2}=\dfrac{3}{5}+2=\dfrac{3}{5}+\dfrac{10}{5}=\dfrac{13}{5}\)}}

a, 23/20

b, 13/72

c, 3/4

d,11/14

e, 13/5

a)\(=\dfrac{12}{20}+\dfrac{11}{20}=\dfrac{23}{20}\)

b)\(=\dfrac{45}{72}-\dfrac{32}{72}=\dfrac{13}{72}\)

c)\(=\dfrac{9\times4}{16\times3}=\dfrac{3}{4}\)

d)\(=\dfrac{4}{7}\times\dfrac{11}{8}=\dfrac{11}{14}\)

e)\(=\dfrac{3}{5}+\dfrac{4}{2}=\dfrac{3}{5}+\dfrac{10}{5}=\dfrac{13}{5}\)

a) \(\dfrac{-5}{6}=\dfrac{-340}{408}\);\(\dfrac{7}{8}=\dfrac{357}{408}\);\(\dfrac{7}{24}=\dfrac{119}{408}\)

\(\dfrac{16}{17}=\dfrac{384}{408}\); \(\dfrac{-3}{4}=\dfrac{-306}{408}\); \(\dfrac{2}{3}=\dfrac{272}{408}\)

Do đó: \(\dfrac{-5}{6}< \dfrac{-3}{4}< \dfrac{7}{24}< \dfrac{2}{3}< \dfrac{7}{8}< \dfrac{16}{17}\)

a) \(\dfrac{3}{11}+\dfrac{14}{11}=\dfrac{17}{11}\)

b) \(\dfrac{1}{16}+\dfrac{3}{4}=\dfrac{1}{16}+\dfrac{12}{16}=\dfrac{13}{16}\)

c) \(\dfrac{2}{20}+\dfrac{7}{10}=\dfrac{2}{20}+\dfrac{14}{20}=\dfrac{16}{20}=\dfrac{4}{5}\)

a) Vì \(\dfrac{x+5}{3}\)= \(\dfrac{x-6}{7}\) nên 7(x+5) = 3(x-6)

=> 7x+ 35 = 3x - 18

7x - 3x = -18 -35

4x = -53

x = -53:4

x = \(\dfrac{-53}{4}\)

\(1,A=-\dfrac{3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)

\(A=-\dfrac{3}{4}.\left(0,125-\dfrac{3}{2}\right):\dfrac{33}{16}-\dfrac{1}{4}\)

\(A=-\dfrac{3}{4}.\left(-\dfrac{11}{8}\right):\dfrac{33}{16}-\dfrac{1}{4}\)

\(A=\dfrac{33}{32}:\dfrac{33}{16}-\dfrac{1}{4}\)

\(A=\dfrac{33}{32}.\dfrac{16}{33}-\dfrac{1}{4}\)

\(A=\dfrac{1}{2}-\dfrac{1}{4}\)

\(A=\dfrac{2}{4}-\dfrac{1}{4}\)

\(A=\dfrac{1}{4}\)

\(D=6\dfrac{5}{12}:2\dfrac{5}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)

\(D=\dfrac{77}{12}:\dfrac{13}{4}+\dfrac{45}{4}.\dfrac{2}{15}\)

\(D=\dfrac{77}{39}+\dfrac{3}{2}\)

\(D=\dfrac{271}{78}\)

\(C=\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)

\(C=\dfrac{5}{16}:0,125-\left(\dfrac{9}{4}-0,6\right).\dfrac{10}{11}\)

\(C=\dfrac{5}{16}:0,125-\dfrac{33}{20}.\dfrac{10}{11}\)

\(C=\dfrac{5}{2}-\dfrac{3}{2}\)

\(C=1\)

\(a,\left(\dfrac{7}{20}+\dfrac{11}{15}-\dfrac{15}{12}\right):\left(\dfrac{11}{20}-\dfrac{26}{45}\right).\)

\(=\left(\dfrac{21}{60}+\dfrac{44}{60}-\dfrac{75}{60}\right):\left(\dfrac{99}{180}-\dfrac{104}{180}\right).\)

\(=\left(\dfrac{65}{60}-\dfrac{75}{60}\right):\left(-\dfrac{5}{180}\right).\)

\(=-\dfrac{10}{60}:\left(-\dfrac{5}{180}\right).\)

\(=-\dfrac{1}{6}:\left(-\dfrac{1}{36}\right).\)

\(=-\dfrac{1}{6}.\left(-36\right).\)

\(=\dfrac{-1.\left(-36\right)}{6}=\dfrac{36}{6}=6.\)

Vậy......

\(b,\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}:\dfrac{15-\dfrac{15}{11}+\dfrac{15}{121}}{16-\dfrac{16}{11}+\dfrac{16}{121}}.\)

\(=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}:\dfrac{15\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}{16\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}.\)

\(=\dfrac{5}{8}:\dfrac{15}{16}.\)

\(=\dfrac{5}{8}.\dfrac{16}{15}=\dfrac{5.16}{8.15}=\dfrac{1.2}{1.3}=\dfrac{2}{3}.\)

Vậy......

c, (làm tương tự câu b).

~ Học tốt!!! ~

a) Ta có: \(\dfrac{5}{8}+\dfrac{3}{17}+\dfrac{4}{18}+\dfrac{20}{-17}+\dfrac{-2}{9}+\dfrac{21}{56}\)

\(=\left(\dfrac{3}{17}-\dfrac{20}{17}\right)+\left(\dfrac{2}{9}-\dfrac{2}{9}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)\)

\(=-1+1=0\)

b) Ta có: \(\left(\dfrac{9}{16}+\dfrac{8}{-27}\right)+\left(1+\dfrac{7}{16}+\dfrac{-19}{27}\right)\)

\(=\left(\dfrac{9}{16}+\dfrac{7}{16}\right)+\left(\dfrac{-8}{27}-\dfrac{19}{27}\right)+1\)

=1-1+1=1