(2^12.3^5-4^6.9^2)/((2^3.3)^6)-5^10.7^3-25^5.49^2/(125.7)^3+5^9.14^3
2^12.3^5-4^6.9^2/(2^2.3)^6+8^4.3^5-5^10.7^3-25^5.49^2/(125.7)^3+5^9.14^3
A=2^12.3^5-4^6.9^2\(2^2.3)^6+8^4.3^5 - 5^10.7^3-25^5.49^2\(125.7)^3+5^9.14^3
(2^12.3^5-4^6.9^2) / [(2^2.3)^6 +8^4.3^5] - (5^10.7^3-25^5.49^2) / [(125.7)^3+5^9.14^3]
Tính;
A= 2^12.3^5-4^6.9^2/(2^2.3)^6 - 5^10.7^3-25^5.49^2/(125.7)^3+5^9.14^3
Ta có: \(\frac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6}\)
\(=\frac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6}=\frac{2^{12}\cdot3^4\left(3-1\right)}{2^{12}\cdot3^6}\)
\(=\frac{3^4\cdot2}{3^6}=\frac{2}{3^2}=\frac29\)
Ta có: \(\frac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
\(=\frac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)
\(=\frac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}=5\cdot\frac{-6}{9}=5\cdot\frac{-2}{3}=-\frac{10}{3}\)
Ta có: \(A=\frac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6}-\frac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
\(=\frac29+\frac{10}{3}=\frac29+\frac{30}{9}=\frac{32}{9}\)
B=2^12.3^5-4^6.9^2/(2^2.3)^6+8^4.3^5-5^10-25^5.49^2/(125.7)^3+5^9.14^3
Tính giá trị biểu thức
P=\(\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^3.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
thực hiện phép tính
a) 81^11.3^17/27^10.9^15
b) A=2^12.3^5-4^6.9^2/(2^2.3^6)+8^4.3^5-5^10.7^3-25^5.49^2/(125.7)^3+5^9
giúp mình vowisddang cần gấp. cảm ơn
a) 81^11.3^17/27^10.9^15
=(9^2)^11.3^17/(3^3)^10.9^15
=3^44.3^17/3^30.3^30
=3^61/3^60
=3
b) A =( (2^12.3^5 - 2^12.3^4)/ (2^12.3^6 + 2^12.3^5) ) - ((5^10.7^3 - 5^10.7^4)/(5^9.7^3 + 5^9.2^3.7^3))
=(2^12.3^4(3-1))/2^12.3^5(3+1) - 5^10.7^3(1-7)/5^9.7^3(1+8)
=2/12- (-30/9)=1/6 + 10/3 = 7/2
\left(4\dfrac{4}{9} + 5\dfrac{4}{11}\right) -1\dfrac{4}{9} =(494+5114)−194
Tính hợp lý:
\(\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\\ =\dfrac{2^{12}\cdot3^5-\left(2^2\right)^6\cdot\left(3^2\right)^2}{2^{12}\cdot3^6}-\dfrac{5^{10}\cdot7^3-\left(5^2\right)^5\cdot\left(7^2\right)^2}{\left(5^3\cdot7\right)^3+5^9\cdot\left(2\cdot7\right)^3}\\ =\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot2^3\cdot7^3}\\ =\dfrac{2^{12}\cdot3^4\left(3-1\right)}{2^{12}\cdot3^6}-\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}\\ =\dfrac{2}{9}-\dfrac{-6}{1+8}=\dfrac{2}{9}+\dfrac{6}{9}=\dfrac{8}{9}\)
\(\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\\ =\dfrac{2^{12}.3^5-2^{12}.3^6}{2^{12}.3^6}-\dfrac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\\ =\dfrac{2^{12}.3^5.\left(1-3\right)}{2^{12}.3^6}-\dfrac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+2^3\right)}\\ =\dfrac{2^{12}.3^5.\left(-2\right)}{2^{12}.3^6}-\dfrac{5^{10}.7^3.\left(-6\right)}{5^9.7^3.9}\\ =\dfrac{-2}{3}-\dfrac{5.\left(-6\right)}{9}\\ =\dfrac{-2}{3}+\dfrac{30}{9}\\ =\dfrac{8}{3}\)
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