Đề bài yêu cầu gì?

thien ty tfboys ơi cách làm là gì

\(B=\frac{3^9-2^3\cdot3^7+2^{10}\cdot3^2-2^{13}}{3^{10}-2^2\cdot3^7+2^{10}\cdot3^3-2^{12}}\)

\(B=\frac{1-2\cdot1+1\cdot1-2}{3-1\cdot1+1\cdot3-1}\)

\(B=\frac{1-2+1-2}{3-1+3-1}\)

\(B=\frac{-1+\left(-1\right)}{2+2}\)

\(B=\frac{-2}{4}\)

\(\Rightarrow B=\frac{-1}{2}\)

Ta có : 

\(A=\frac{3^9-2^3.3^7+2^{10}.3^2-2^{13}}{3^{10}-2^2.3^7+2^{10}.3^3-2^{12}}\)

\(A=\frac{3^7\left(3^2-2^3\right)+2^{10}\left(3^2-2^3\right)}{3^7\left(3^3-2^2\right)+2^{10}\left(3^3-2^2\right)}\)

\(A=\frac{\left(3^2-2^3\right)\left(3^7+2^{10}\right)}{\left(3^3-2^2\right)\left(3^7+2^{10}\right)}\)

\(A=\frac{3^2-2^3}{3^3-2^2}\)

\(A=\frac{9-8}{27-4}\)

\(A=\frac{1}{23}\)

Vậy \(A=\frac{1}{23}\)

Chúc bạn học tốt ~ 

\(\frac{-11^5.13^7}{11^5.13^8}=\frac{\left(-1\right).11^5.13^7}{11^5.13^7.13}=\frac{-1}{13}\)

\(\frac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\frac{2^{10}.\left(3^{10}-3^9\right)}{2^9.3^{10}}=\frac{2^9.2.3^9.\left(3-1\right)}{2^9.3.3^9}=\frac{2.2}{3}=\frac{4}{3}\)

a, \(\frac{-11^5.13^7}{11^5.13^8}=\frac{-1}{13}\)

b, \(\frac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\frac{2^{10}.3^9.\left(1-3\right)}{2^9.3^{10}}=\frac{-4}{3}\)

a: Ta có: \(3^{2020}=3^{2018}\cdot3^2=3^{2018}\cdot9\)

mà 9<10

nên \(3^{2020}< 10\cdot3^{2018}\)

(x-1)^2:5^21=25^30.5

(x-1)^2=25^30.5.5^21

           =(5^2)^30.5^22

           = 5^60.5^22

(x-1)^2 =5^82

(x-1)^2=(5^41)^2

x-1=5^41

x=5^41+1

2.3^x+1=10.3^12+8:3^12

2.3^x+1=10+8=18

3^x+1=18/2=9

3^x+1=3^2

x+1=2

x=1