giúp mk ik, năn nỉ đó.

\(=\dfrac{2^{19}\cdot3^9-3\cdot3^8\cdot2^{18}\cdot5}{2^{19}\cdot3^9+2^{20}\cdot3^{10}}=\dfrac{-3^{10}\cdot2^{18}}{2^{19}\cdot3^9\cdot7}=-\dfrac{3}{14}\)

a) Ta có : 2 + 4 + 6 + 8 +... + 16 + 18

SSH : (18 - 2) : 2 + 1 = 9(số)

Tổng : (2 + 18).9 : 2 = 90

\(A=\frac{27\cdot45+27\cdot45}{90}=\frac{27\left(45+45\right)}{90}=\frac{27\cdot90}{90}=27\)

b) Ta có : 135.1420 + 45.780.3 = 135.1420 + 135.780 = 135(1420 + 780) = 135.2200 (*)

3 + 6 + 9 + 12 + ... + 24 + 27

Số số hạng : (27 - 3) : 3 + 1 = 9(số)

Tổng : (3 + 27).9 : 2 = 135 (**)

Từ (*) và (**) suy ra 

=> \(B=\frac{135\cdot2200}{135}=2200\)

a)           

2 + 4 + 6 + ... + 16 + 18                      

Số số hạng : 

( 18 - 2 ) / 2 + 1 = 9 

Tổng : 

( 18 + 2 ) x 9 / 2 = 90 

\(A=\frac{27\cdot45+27\cdot45}{90}\) 

\(=\frac{27\left(45+45\right)}{90}\) 

\(=\frac{27\cdot90}{90}\)    

\(=27\) 

b) 

3 + 6 + 9 + 12 + ... + 24 + 27 

Số số hạng : 

( 27 - 3 ) / 3 + 1 = 9 

Tổng : 

( 27 + 3 ) x 9 /2 = 135 

\(B=\frac{135\cdot1420+45\cdot780\cdot3}{135}\)  

\(=\frac{135\cdot1420+135\cdot780}{135}\)      

\(=\frac{135\left(1420+780\right)}{135}\)                

\(=\frac{135\cdot2200}{135}\)    

= 2200 

\(\frac{3^7.16^3}{12^5.27^2}=\frac{3^7.\left(2^4\right)^3}{\left(2^2.3\right)^5.\left(3^3\right)^2}=\frac{3^7.2^{12}}{2^{10}.3^5.3^6}=\frac{3^7.2^{12}}{2^{10}.3^{11}}=\frac{2^2}{3^4}=\frac{4}{81}\)

\(\frac{3^7.16^3}{12^5.27^2}=\frac{3^7.\left(2^4\right)^3}{\left(2^2.3\right)^5.\left(3^3\right)^2}=\frac{3^7.2^{12}}{2^{10}.3^{11}}=\frac{2^2}{3^4}\)

56/7*2 = 4
12*3+6 = 42
27/ 3/ 3= 3
2*6/2 = 6
24/3/2 = 4

\(\frac{2^{19}.27^3+15^4.4^9.9^4}{6^9.2^{10}+12^{10}}\)

\(=\frac{2^{19}.\left(3^3\right)^3+\left(3.5\right)^4.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)

\(=\frac{2^{19}.3^9+3^{12}.5^4.2^{18}}{2^{19}.3^9+2^{20}.3^{10}}\)

\(=\frac{2^{18}.3^9.\left(2+3^3+5^4\right)}{2^{19}.3^9.\left(1+2+3\right)}\)

\(=\frac{654}{2.6}\)

\(=\frac{109}{2}\)

Chúc bn học tốt !!!!

Ta có:

\(\frac{2^{20}\cdot27^3+30\cdot4^9\cdot9^4}{6^9\cdot4^5+12^{10}}=\frac{2^{20}\cdot\left[3^3\right]^3+2\cdot3\cdot5\cdot\left[2^2\right]^9\cdot\left[3^2\right]^4}{2^9\cdot3^9\cdot\left[2^2\right]^5+3^{10}\cdot\left[2^2\right]^{10}}=\frac{2^{20}\cdot3^{3\cdot3}+2\cdot3\cdot5\cdot2^{2\cdot9}\cdot3^{2\cdot4}}{2^9\cdot3^9\cdot2^{2\cdot5}+3^{10}\cdot2^{2\cdot10}}\)

\(=\frac{2^{20}\cdot3^9+2\cdot3\cdot5\cdot2^{18}\cdot3^8}{2^9\cdot3^9\cdot2^{10}+3^{10}\cdot2^{20}}=\frac{2^{20}\cdot3^9+2^{19}\cdot3^9\cdot5}{2^{19}\cdot3^9+3^{10}\cdot2^{20}}=\frac{2^{19}\cdot3^9\left[2+5\right]}{2^{19}\cdot3^9\left[1+3\cdot2\right]}=\frac{2+5}{1+6}=\frac{7}{7}=1\)

\(\begin{array}{l}a)\frac{{{3^{12}} + {3^{15}}}}{{1 + {3^3}}}\\ = \frac{{{3^{12}} + {3^{12}}{{.3}^3}}}{{1 + {3^3}}}\\ = \frac{{{3^{12}}.(1 + {3^3})}}{{1 + {3^3}}}\\ = {3^{12}}\\b)2:{\left( {\frac{1}{2} - \frac{2}{3}} \right)^2} + 0,{125^3}{.8^3} - {( - 12)^4}:{6^4}\\ = 2:{\left( {\frac{3}{6} - \frac{4}{6}} \right)^2} + {(0,125.8)^3} - {12^4}:{6^4}\\ = 2:{\left( {\frac{{ - 1}}{6}} \right)^2} + {1^3} - {(\frac{{12}}{6})^4}\\ = 2:\frac{1}{{36}} + 1 - {2^4}\\ = 2.36 + 1 - 16\\ = 72 + 1 - 16=57\end{array}\)