a) \(A=27\cdot36+73\cdot99+27\cdot14-49\cdot73\)

\(A=27\cdot\left(36+14\right)+73\cdot\left(99-49\right)\)

\(A=27\cdot50+73\cdot50\)

\(A=50\cdot\left(27+73\right)\)

\(A=50\cdot100\)

\(A=5000\)

b) \(B=\left(4^5\cdot10\cdot5^6+25^5\cdot2^8\right):\left(2^8\cdot5^4+5^7\cdot2^5\right)\)

\(B=\dfrac{\left(2^2\right)^5\cdot2\cdot5\cdot5^6+\left(5^2\right)^5\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}\)

\(B=\dfrac{2^{11}\cdot5^7+5^{10}\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}\)

\(B-\dfrac{2^8\cdot5^7\cdot\left(2^3\cdot1+5^3\cdot1\right)}{2^5\cdot5^4\cdot\left(2^3\cdot1+5^3\cdot1\right)}\)

\(B=\dfrac{2^8\cdot5^7}{2^5\cdot5^4}\)

\(B=2^3\cdot5^3\)

\(B=10^3\)

\(B=1000\)

2:

a: \(=\dfrac{1}{3}\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)=-\dfrac{1}{3}\cdot2=-\dfrac{2}{3}\)

1:

\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)

\(=-4-\dfrac{1}{4}=-\dfrac{17}{4}\)

Bài 1:

\(A=\left(7-\dfrac{3}{4}+\dfrac{1}{3}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\)

\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)

\(A=\left(7-6-5\right)-\left(\dfrac{3}{4}+\dfrac{5}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{3}+\dfrac{4}{3}-\dfrac{5}{3}\right)\)

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

\(A=-4-\dfrac{1}{4}+\dfrac{0}{3}\)

\(A=-\dfrac{16}{4}-\dfrac{1}{4}+0\)

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

\(A=-\dfrac{17}{4}\)

Bài 2:

\(\dfrac{1}{3}\cdot-\dfrac{4}{5}+\dfrac{1}{3}\cdot-\dfrac{6}{5}\)

\(=\dfrac{1}{3}\cdot\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{-4-6}{5}\)

\(=\dfrac{1}{3}\cdot\dfrac{-10}{5}\)

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

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

a: =382-282+531-331

=100+200=300

b: =(7-8)+(9-10)+...+(2009-2010)

=(-1)+(-1)+....+(-1)

=-1002

c: =-(1+2+3+...+2009+2010)

=-2010*2011/2=-2021055

a: \(A=\dfrac{5}{7}-\dfrac{2}{7}+\dfrac{8}{11}+\dfrac{3}{11}+\dfrac{1}{2}=\dfrac{3}{7}+\dfrac{1}{2}+1=\dfrac{6+7+14}{14}=\dfrac{27}{14}\)

b: \(B=\dfrac{11}{17}+\dfrac{6}{17}-\dfrac{8}{19}-\dfrac{30}{19}+\dfrac{-3}{4}=1-2-\dfrac{3}{4}=-1-\dfrac{3}{4}=-\dfrac{7}{4}\)

c: \(C=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}=\dfrac{49}{50}\)

a, \(\dfrac{3}{5}.\left(-\dfrac{8}{3}\right)-\dfrac{3}{5}:\left(-\dfrac{3}{2}\right)=\dfrac{3}{5}.\left(-\dfrac{8}{3}\right)-\dfrac{3}{5}.\left(-\dfrac{2}{3}\right)==\dfrac{3}{5}\left(-\dfrac{8}{3}-\dfrac{2}{3}\right)=\dfrac{3}{5}.\left(-\dfrac{10}{3}\right)=-2\)

b, \(-\dfrac{5}{6}.\left(-\dfrac{12}{7}\right)-\left(-\dfrac{21}{15}\right)=-\dfrac{5}{6}.\left(-\dfrac{12}{7}\right)+\dfrac{7}{5}=\dfrac{10}{7}+\dfrac{7}{5}=\dfrac{50+49}{35}=\dfrac{99}{35}\)

a: \(=\dfrac{3}{5}\cdot\left(-\dfrac{8}{3}+\dfrac{-2}{3}\right)=\dfrac{3}{5}\cdot\dfrac{-10}{3}=-2\)

c: \(=\left(0.125\right)^{650}\cdot8^{102}\)

\(=\left(0.125\cdot8\right)^{102}\cdot\left(0.125\right)^{548}\)

\(=\dfrac{1}{8^{548}}\)

a: =46-16+35-5=30+30=60

b: =32-12+34-14+36-16+38-18-10

=20+20+20+20-10

=80-10=70

c: \(=125-125-170+120=-50\)

d: =(-1)+(-1)+...+(-1)

=-50

#\(N\)

`a, 4573 + 46 - 4573 + 35 - 16 - 5`

`= (4573-4573) + (46 - 16)+(35-5)`

`= 0 +30+30 = 60`

`b, 32+34+36+38-10-12-14-16-18`

`= (32-12)+(34-14)+(38-18)+10`

`= 20+20+20+10 = 70`

`c, 125-170+120+(-125)`

`= (125 + -125)-170+120`

`= 0-170+120`

`=-170 + 120 = -50`

`d, 1-2+3-4+...-98+99-100`

Các số hạng có trong biểu thức: \(\left(100-1\right)\div1+1=100\) `(` số hạng `)`

`=> (1-2)+(3-4)+...+(97-98)+(99-100)`

Các cặp mà trong bthuc có là: \(100\div2=50\) 

`=> (-1)+(-1)+...(-1)+(-1) = (-50)`

`e,`

*Mình xp sửa đề phải là `1-5 + 7-11+...+997-1001` nhỉ? Vì để như vậy nó lẻ á ._.

`1-5+7-11+...+997-1001`

Các số hạng có trong biểu thức là: \(\left(1001-1\right)\div4+1=251\) `(` số `)`

`-> (1-5)+(7-11)+...+(997-1001)`

Các cặp được ghép ở trong bthuc là: \(251\div2=125,5\) 

`-> (-4)+(-4)+...+(-4) = (-4)*125,5 = -502`

2:

a: x=2,4-0,4=2

b: =>2x=-1,5+0,8=-0,7

=>x=-0,35

c: =>x-16=-15

=>x=1

3/5

3/5

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

Bài 1: 

a: \(\dfrac{25}{42}-\dfrac{20}{63}=\dfrac{75-40}{126}=\dfrac{35}{126}=\dfrac{5}{18}\)

b: \(\dfrac{9}{20}-\dfrac{13}{75}-\dfrac{1}{6}=\dfrac{135}{300}-\dfrac{52}{300}-\dfrac{50}{300}=\dfrac{33}{300}=\dfrac{11}{100}\)