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

=\(\left(3.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)\right):\left(11\left(\dfrac{1}{7}+\dfrac{1}{13}+\dfrac{1}{4}-\dfrac{1}{5}\right)\right)\)

= \(\dfrac{3}{11}\)

Mình sửa số chia thành \(\left(\dfrac{11}{7}+\dfrac{11}{13}+2,75-2,2\right)\)

a, \(\dfrac{0,75-0,6+\dfrac{3}{7}+\dfrac{3}{13}}{2,72-2,2+\dfrac{11}{7}+\dfrac{11}{13}}\)

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

= \(\dfrac{3.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}\)

= \(\dfrac{3}{11}\)

b. \(\dfrac{0,357-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{0,625-0,5+\dfrac{5}{11}+\dfrac{5}{12}}\)

= \(\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{\dfrac{5}{8}-\dfrac{5}{10}+\dfrac{5}{11}+\dfrac{5}{12}}\)

= \(\dfrac{3.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{5.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}\)

= \(\dfrac{3}{5}\)

c, \(-\left|-1,5\right|.\left(1\dfrac{1}{3}-2\right)-\left|-\dfrac{2}{3}\right|\)

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

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

= \(1-\dfrac{2}{3}=\dfrac{1}{3}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(C=1\)

ngu như con bò tót, ko biết 1+1=2.

5) \(\left(-2\right)^2+\sqrt{36}-\sqrt{9}+\sqrt{25}\)

=\(4+6-3+5\)

=\(12\)

2) \(\dfrac{11}{25}.\left(-24,8\right)-\dfrac{11}{25}.75,2\)

=\(\dfrac{11}{25}.\left(-24,8-75,2\right)\)

=\(\dfrac{11}{25}.\left(-100\right)\)

=\(-44\)

\(6\)) \(\left(-\dfrac{1}{3}\right)^2\).\(\dfrac{4}{11}+\dfrac{7}{11}.\left(-\dfrac{1}{3}\right)^2\)

=\(\left(-\dfrac{1}{3}\right)^2.\left(\dfrac{4}{11}+\dfrac{7}{11}\right)\)

=\(\left(-\dfrac{1}{3}\right)^2.1\)

=\(\dfrac{1}{9}\)

A= \(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{9}{11}=\dfrac{1}{3}-\dfrac{7}{9}=\dfrac{3}{9}-\dfrac{7}{9}=-\dfrac{4}{9}\)

\(B=\left(\dfrac{1}{5}+\dfrac{2}{15}+\dfrac{2}{3}\right)+\left(-\dfrac{2}{7}+\dfrac{1}{42}-\dfrac{13}{28}-\dfrac{1}{4}\right)\)

\(=\dfrac{3+2+10}{15}+\dfrac{-2\cdot12+2-13\cdot3-21}{84}\)

=1-82/84

=2/84=1/42

\(C=\dfrac{1}{50}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\right)\)

\(=\dfrac{1}{50}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)

\(=\dfrac{1}{50}-1+\dfrac{1}{50}=\dfrac{1}{25}-1=-\dfrac{24}{25}\)

\(D=\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{3}{11}\)

Mấy bài này bạn tự làm đi, chuyển vế tìm x gần giống cấp I mà.

b)\(\dfrac{-3}{5}.x=\dfrac{1}{4}+0,75\)

=>\(\dfrac{-3}{5}.x=1\)

=>\(x=1:\dfrac{-3}{5}\)

=>\(x=\dfrac{-5}{3}\)

Vậy \(x=\dfrac{-5}{3}\)

a: \(A=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)

b: \(B=\dfrac{8+5}{10}:\dfrac{-5}{13}=\dfrac{13}{10}\cdot\dfrac{13}{-5}=-\dfrac{169}{100}\)

c: \(C=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}+\dfrac{132}{132}-\dfrac{84}{132}\right)\)

\(=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)