a, \(A\left(x\right)=x^2-2x^2+x^2+x+2=x+2\)

\(B\left(x\right)=2x^3-3x^2+x-2x^2+3=2x^3-5x^2+x+3\)

b, \(A\left(x\right)+B\left(x\right)=2x^3-5x^2+x+3+x+2=2x^3-5x^2+2x+5\)

\(\left(\dfrac{-9}{24}-\dfrac{6}{24}+\dfrac{10}{24}\right)\times\dfrac{3}{2}\)

\(=\left(\dfrac{-5}{24}\right)\times\dfrac{3}{2}\)

\(=\dfrac{-5}{16}\)

\(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\)

\(=\dfrac{5}{10}+\dfrac{8}{10}\)

\(=\dfrac{13}{10}\)

\(\dfrac{1}{\dfrac{13}{15}}\times\dfrac{3}{4}-\left(\dfrac{11}{20}-\dfrac{5}{20}\right)\times\dfrac{7}{5}\)

\(=\dfrac{15}{13}\times\dfrac{3}{4}-\dfrac{3}{10}\times\dfrac{7}{5}\)

\(=\dfrac{45}{52}-\dfrac{21}{50}\)

\(=\dfrac{1125}{1300}-\dfrac{546}{1300}\)

\(=\dfrac{579}{1300}\)

\(5E=5^2+5^3+...+5^{501}\)

\(5E-E=5^{501}-5\)

\(4E=5^{501}-5\)

\(E=\dfrac{5^{501}-5}{4}\)

Số đội nhiều nhất chia được là ước chung lớn nhất của 120, 1008 và 1680
Ta có:
\(120=2^3.3.5\)
\(1008=2^4.3^2.7\)
\(1680=2^4.3.5.7\)
Suy ra: \(ƯCLN\left(120,1008,1680\right)=2^3.3=8.3=24\)
\(\Rightarrow\) Có thể lập nhiều nhất 24 đội phản ứng nhanh trong đó bác sĩ, y tá, bộ đội của mỗi đội bằng nhau
Vậy...

x + 30 = -50

=> x = -50 - 30

=> x = -80  

a; A = \(\dfrac{3^{10}.11+3^{10}.5}{3^9.2^4}\) = \(\dfrac{3^{10}.\left(11+5\right)}{3^9.2^4}\) = \(\dfrac{3^{10}.16}{3^9.16}\) = 3

b; B = \(\dfrac{2^{10}.13+2^{10}.65}{2^8.104}\) 

= \(\dfrac{2^{10}.\left(13+65\right)}{2^8.104}\) 

= \(\dfrac{2^{10}.78}{2^8.2^2.26}\) 

= \(\dfrac{2^{10}.26.3}{2^{10}.26}\) 

= 3 

b; B = \(\dfrac{2^{10}.13+3^{10}.5}{3^9.2^4}\)

    B =  \(\dfrac{3^{10}.\left(13+5\right)}{3^9.2^4}\)

    B =  \(\dfrac{3^{10}.18}{3^9.2^4}\) 

    B =  \(\dfrac{3^{10}.3^2.2}{3^9.2^4}\)

    B =   \(\dfrac{3^{12}.2}{3^9.2^4}\)

    B =    \(\dfrac{3^3}{2^4}\) 

    B = \(\dfrac{27}{8}\)

a: \(4^{10}\cdot2^{30}=2^{20}\cdot2^{30}=2^{50}\)

b: \(9^{25}\cdot27^4\cdot81^3=3^{50}\cdot3^{12}\cdot3^{12}=3^{74}\)

c: \(25^{50}\cdot125^5=\left(5^2\right)^{50}\cdot\left(5^3\right)^5=5^{115}\)

d: \(64^3\cdot4^8\cdot16^4=\left(4^3\right)^3\cdot4^8\cdot\left(4^2\right)^4=4^9\cdot4^8\cdot4^8=4^{25}\)

e: \(3^8:3^6=3^{8-6}=3^2\)

f: \(2^{10}:8^3=2^{10}:2^9=2\)

g: \(12^7:6^7=\left(\dfrac{12}{6}\right)^7=2^7\)

h: \(21^5:81^3=\dfrac{7^5\cdot3^5}{3^3\cdot27^3}=\dfrac{7^5}{27^3}\)

i: \(4^9:64^2=4^9:\left(4^3\right)^2=4^9:4^6=4^3\)

j: \(2^{25}:32^4=2^{25}:2^{20}=2^5\)

k: \(125^3:25^4=\left(5^3\right)^3:\left(5^2\right)^4=5^9:5^8=5\)

\(A=1+2+2^2+...+2^{99}\\ 2A=2+2^2+2^3+...+2^{100}\\ 2A-A=\left(2+2^2+2^3+...+2^{100}\right)-\left(1+2+2^2+...+2^{99}\right)\\ A=2^{100}-1\)

\(=>A+1=2^{100}-1+1=2^{100}\)

Mà: \(A+1=2^n=>2^n=2^{100}\)

\(=>n=100\)

Có :

9 - x = 15

     x = 9 - 15

     x = - 6

Mà -6 \(\notin\) N ⇒ D = { }

9 - x = 15

=> x = 9 - 15 

=> x = -6

Mà x ∈ N => K có x thỏa mãn 

=> D = ∅

\(170=17\cdot2\cdot5;290=29\cdot2\cdot5\)

=>\(BCNN\left(170;290\right)=17\cdot29\cdot2\cdot5=4930\)

\(a⋮170;a⋮290\)

=>\(a\in BC\left(170;290\right)\)

mà a nhỏ nhất

nên a=BCNN(170;290)

=>a=4930