a) (-125) . 4

= -500

b) (-2017) . (-2016) . 0

= 0

c) (-12) . (-11)

= 132

d) (-4) . (-8) . (-25) . (-125)

= [(-4).(-25)].[(-8).(-125)]

= 100.1000

= 100000

e) (-2).(-5).(-7)

= [(-2).(-5)].(-7)

= 10.(-7)

= -70

a: \(\left(-125\right)\cdot4=-\left(125\cdot4\right)=-500\)

b: \(\left(-2017\right)\cdot\left(-2016\right)\cdot0=0\cdot2016\cdot2017=0\)

c: \(\left(-12\right)\cdot\left(-11\right)=12\cdot11=12\cdot10+12=120+12=132\)

d: \(\left(-4\right)\cdot\left(-8\right)\cdot\left(-25\right)\cdot\left(-125\right)\)

\(=4\cdot8\cdot25\cdot125\)

\(=4\cdot25\cdot8\cdot125\)

\(=100\cdot1000=100000\)

e: \(\left(-2\right)\cdot\left(-5\right)\cdot\left(-7\right)\)

\(=-7\cdot2\cdot5\)

\(=-7\cdot10=-70\)

\(\dfrac{5}{7}\).\(\dfrac{-3}{11}\)-\(\dfrac{5}{7}\).\(\dfrac{8}{11}\)+1\(\dfrac{5}{7}\)

= \(\dfrac{5}{7}\).(\(\dfrac{-3}{11}\) - \(\dfrac{8}{11}\)) + 1\(\dfrac{5}{7}\)

= \(\dfrac{5}{7}\). -1             + 1\(\dfrac{5}{7}\)

= \(\dfrac{-5}{7}\)               +1\(\dfrac{5}{7}\)

= -1

\(\dfrac{5}{7}.-\dfrac{3}{11}-\dfrac{5}{7}.\dfrac{8}{11}+1\dfrac{5}{7}\)

\(=\dfrac{5}{7}.\left(\dfrac{3}{11}-\dfrac{8}{11}\right)+1\dfrac{5}{7}\)

\(=\dfrac{5}{7}.\left(-1\right)+1\dfrac{5}{7}\)

\(=1\)

\(12,25+\left(-21,37\right)-\left(14,67-21,75\right)\)

\(=12,25-21,37-14,67+21,75\)

\(=12,25+21,75-21,37-14,63-0,04\)

\(=34-36-0,04\)

= -2-0,04=-2,04

120:{54-[50:2.(3^2-2.4)]}

= 120:{ 54-[25.(9-8)]}

= 120:{54-[25.1]}

= 120:{54-25}

= 120:29

=\(\dfrac{120}{29}\)

e: \(=\dfrac{2^{10}\cdot3^9-2^9\cdot3^8}{2^{10}\cdot3^8}=\dfrac{2^9\cdot3^8\left(2\cdot3-1\right)}{2^{10}\cdot3^8}=\dfrac{5}{2}\)

g: \(=125+\dfrac{3}{5}-97-\dfrac{2}{3}+97+\dfrac{2}{5}-125-\dfrac{1}{3}=0\)

\(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{323}\\ =\dfrac{1}{2}\times\left(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+...+\dfrac{2}{17\times19}\right)\\=\dfrac{1}{2}\times\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{17}-\dfrac{1}{19}\right)\\ =\dfrac{1}{2}\times\left(\dfrac{1}{1}-\dfrac{1}{19}\right)\\ =\dfrac{1}{2}\times\dfrac{18}{19}=\dfrac{9}{19}\)

đề bài đây: 

\(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{323}\)

43.11 = (10 + 10 + 10 + 10 + 3).11 = 110 + 110 + 110 + 110 + 33 = 440 + 33 = 473

67.99 = 67. (100 - 1) = 6 700 - 67 = 6 633

chắc làm theo cách này: )