\(\frac{2^{12}.3^5-\left(2^2\right)^6.3^6}{2^{12}.\left(3^2\right)^3+\left(2^3\right)^4.3^3}\)
\(\frac{2^{12}.3^5.\left(1-3^{ }\right)}{2^{12}.3^3.\left(3^3-1\right)}\)
\(\frac{2^{12}.3^5.\left(-2\right)}{2^{12}.3^3.8}\)
\(\frac{3^2.\left(-1\right)}{4}\)
\(\frac{-9}{4}\)
VẬy.......................
nhớ tk cho mình nha
\(TínhA=\frac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\frac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
\(B=\frac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\frac{56^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
Tính A:\(\frac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\frac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
Rút gọn :
P=\(\frac{2^{12}\cdot3^5-4^6\cdot3^6}{2^{12}\cdot9^3-8^4\cdot3^5}\)
thực hiện phép tính:
A = \(\frac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\frac{5^{10}\cdot7^3-25^5\cdot49}{\left(125\cdot7\right)^3+5^9\cdot14^3}^2\)
tính A=\(y=\frac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\frac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
thankS trc nhé =))~~~~
\(\frac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\frac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
giúp mk zới nhá vế sau mk chịu òi
B =\(\frac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}\) - \(\frac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9+14^3}\)
A = \(\frac{2^{12}\cdot3^2-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4+3^5}-\frac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
