Question

mn giúp e bài 40 với ạ , e cần gấp 

Answer 1

Bài 40:

1: \(A=\frac{35\cdot\left(27^8+2\cdot9^{11}\right)}{15\left(81^6-12\cdot3^{19}\right)}\)

\(=\frac73\cdot\frac{3^{12}+2\cdot3^{22}}{3^{24}-2^2\cdot3\cdot3^{19}}=\frac73\cdot\frac{3^{12}\left(1+2\cdot3^{10}\right)}{3^{20}\left(3^4-2^2\right)}\)

\(=\frac73\cdot\frac{1}{3^8}\cdot\frac{1+2\cdot3^{10}}{81-4}=\frac{7}{3^9}\cdot\frac{118099}{77}=\frac{118099}{11\cdot3^9}\)

2: \(B=\frac{\left(n+2\right)!+\left(n+3\right)!}{\left(n+2\right)!-\left(n+3\right)!}\)

\(=\frac{\left(n+2\right)!\cdot\left\lbrack1+\left(n+3\right)\right\rbrack}{\left(n+2\right)!\cdot\left\lbrack1-\left(n+3\right)\right\rbrack}\)

\(=\frac{1+\left(n+3\right)}{1-\left(n+3\right)}=\frac{1+n+3}{1-n-3}=\frac{n+4}{-n-2}\)