a,

\(\dfrac{4^6\cdot9^5+6^9\cdot120}{-8^4\cdot3^{12}-6^{11}}=\dfrac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{-2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{-2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{-2^{11}\cdot3^{11}\left(2\cdot3-1\right)}=\dfrac{2^{13}\cdot3^{11}}{-2^{11}\cdot3^{11}\left(2\cdot3+1\right)}=\dfrac{2^2}{7}=\dfrac{4}{7}\)

b,

\(\dfrac{1}{1-\dfrac{1}{1-2-1}}+\dfrac{1}{1+\dfrac{1}{1+2-1}}=\dfrac{1}{1-\dfrac{1}{-2}}+\dfrac{1}{1+\dfrac{1}{2}}=\dfrac{1}{1+\dfrac{1}{2}}+\dfrac{1}{1+\dfrac{1}{2}}=\dfrac{2}{\dfrac{3}{2}}=\dfrac{4}{3}\)

\(a,\frac{-7}{25}.\frac{11}{13}+\frac{-7}{25}.\frac{2}{13}-\frac{18}{25}\)

\(=\frac{-7}{25}.\left(\frac{11}{13}+\frac{2}{13}\right)-\frac{18}{25}=\frac{-7}{25}-\frac{18}{25}=-1\)

\(b,\frac{5}{7}.\frac{1}{3}-\frac{5}{7}.\frac{1}{4}-\frac{5}{7}.\frac{1}{12}=\frac{5}{7}.\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)=\frac{5}{7}.\left(\frac{4}{12}-\frac{3}{12}-\frac{1}{12}\right)\)

\(=\frac{5}{7}.0=0\)

c)\(5\frac{2}{5}.4\frac{2}{7}+5\frac{5}{7}.5\frac{2}{5}=\frac{27}{5}.\frac{30}{7}+\frac{40}{7}.\frac{27}{5}=\frac{27}{5}.\left(\frac{30}{7}+\frac{40}{7}\right)\)

\(=\frac{27}{5}.10=27.2=54\)

\(d,75\%-1\frac{1}{2}+0,5:\frac{5}{12}-\left(\frac{-1}{2}\right)^2=\frac{3}{4}-\frac{3}{2}+\frac{1}{2}.\frac{12}{5}-\frac{1}{4}\)

\(=\left(\frac{3}{4}-\frac{1}{4}\right)-\frac{3}{2}+\frac{6}{5}=\frac{1}{2}-\frac{3}{2}+\frac{6}{5}=-1+\frac{6}{5}=\frac{-5}{5}+\frac{6}{5}=\frac{1}{5}\)

cho mk hỏi, bn có thể lm thêm c d dc ko

Ta có: \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)....\left(1-\frac{1}{780}\right)\)

\(=\frac{2}{3}.\frac{5}{6}...\frac{779}{780}\)

\(=\frac{4}{6}.\frac{10}{12}....\frac{1558}{1560}\)

\(=\frac{1.4.2.5....38.41}{2.3.3.4....39.40}=\frac{\left(1.2.3..38\right)\left(4.5...41\right)}{\left(2.3.4...39\right)\left(3...40\right)}=\frac{41}{39.3}=\frac{41}{117}\)

 \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)........\left(1-\frac{1}{780}\right)\)

\(=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}........\frac{779}{780}\)

\(=\frac{4}{6}.\frac{10}{12}\frac{18}{20}.\frac{28}{30}.........\frac{1558}{1560}\)

\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}...............\frac{38.41}{39.40}\)

\(=\frac{\left(1.2.3.4......38\right)\left(4.5.6.7..........41\right)}{\left(2.3.4.5.........39\right)\left(3.4.5.6.........40\right)}\)

\(=\frac{1.41}{39.3}\)

\(=\frac{41}{117}\)

Vậy \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)........\left(1-\frac{1}{780}\right)=\frac{41}{117}\)

\(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{780}\right)\)

\(=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}...\frac{779}{780}=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}...\frac{1558}{1560}\)

\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{38.41}{39.40}\)

\(=\frac{\left(1.2.3...38\right)\left(4.5.6...61\right)}{\left(2.3.4...39\right)\left(3.4.5...40\right)}=\frac{1.41}{39.3}=\frac{41}{117}\)

\(=1+\frac{1}{1+\frac{1}{1+\frac{1}{\frac{3}{2}}}}=1+\frac{1}{1+\frac{1}{1+\frac{2}{3}}}=1+\frac{1}{1+\frac{1}{\frac{5}{3}}}=1+\frac{1}{1+\frac{3}{5}}=1+\frac{1}{\frac{8}{5}}=1+\frac{5}{8}=\frac{13}{8}\)

a,(1/3/7-2/1/4) . 3/1/3

=     -23/28     .3/1/3

=               -115/42

b,(2/1/3+3/1/2):(-4/1/6+3/1/7)+7/1/2

=        35/6     :       -43/42    +7/1/2

=                  -245/43            +7/1/2

=                               155/86

\(\left(1\frac{3}{7}-2\frac{1}{4}\right).3\frac{1}{3}\)

\(=\left(\frac{10}{7}-\frac{9}{4}\right).\frac{10}{3}\)

\(=-\frac{23}{28}.\frac{10}{3}\)

\(=\frac{-115}{42}\)

a,\(\left(1\frac{3}{7}-2\frac{1}{4}\right).3\frac{1}{3}=\frac{-23}{28}.\frac{10}{3}=\frac{-115}{42}\)

Còn lại bạn tự giải nha sử dụng tính chất giao hoán kết hợp để tính nha

\(\frac{5}{a}+\frac{3}{a+4}=\frac{5.\left(a+4\right)+3a}{a.\left(a+4\right)}=\frac{5a+20+3a}{a^2+4a}\)

                          \(=\frac{8a+20}{a^2+4a}\)

\(\frac{4}{c-5}+\frac{2}{2c+3}\) \(=\frac{4\left(2c+3\right)+2\left(c-5\right)}{\left(c-5\right)\left(2c+3\right)}\)

                                        \(=\frac{8c+12+2c-10}{2c^2+3c-10c-15}\)

                                         \(=\frac{10c-2}{2c^2-7c-15}\)

câu còn lại tương tự nha

mk phải đi học rồi