Giải thích các bước giải:
a.Ta có:
$A=\dfrac{9^3\cdot 7-27^2\cdot 3}{3^4\cdot 2+9^2\cdot 5^2}$
$\to A=\dfrac{(3^2)^3\cdot 7-(3^3)^2\cdot 3}{3^4\cdot 2+(3^2)^2\cdot 5^2}$
$\to A=\dfrac{3^{2\cdot 3}\cdot 7-3^{3\cdot 2}\cdot 3}{3^4\cdot 2+3^{2\cdot 2}\cdot 5^2}$
$\to A=\dfrac{3^{6}\cdot 7-3^{6}\cdot 3}{3^4\cdot 2+3^{4}\cdot 5^2}$
$\to A=\dfrac{3^{6}\cdot (7-3)}{3^4\cdot (2+5^2)}$
$\to A=\dfrac{3^{6}\cdot 4}{3^4\cdot 27}$
$\to A=\dfrac{3^{6}\cdot 4}{3^4\cdot 3^3}$
$\to A=\dfrac{3^{6}\cdot 4}{3^{4+3}}$
$\to A=\dfrac{3^{6}\cdot 4}{3^{7}}$
$\to A=\dfrac{4}{3}$
b.Ta có:
$B=\dfrac{3^2\cdot 64^2-12^2\cdot 16^2\cdot 19^0}{3+6+9+...+96+99}$
$\to B=\dfrac{3^2\cdot (2^6)^2-(3\cdot 2^2)^2\cdot (2^4)^2\cdot 1}{3(1+2+3+...+32+33)}$
$\to B=\dfrac{3^2\cdot 2^{12}-3^2\cdot 2^4\cdot 2^8}{3\cdot \dfrac{33\cdot (33+1)}{2}}$
$\to B=\dfrac{3^2\cdot 2^{12}-3^2\cdot 2^{12}}{3\cdot \dfrac{33\cdot (33+1)}{2}}$
$\to B=\dfrac{0}{3\cdot \dfrac{33\cdot (33+1)}{2}}$
$\to B=0$
