Question

Tính A :

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

Giúp mình với!!!!!!!!

Answer 1

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 ~