Thực hiện phép tính A=(2^12.3^5-4^6.9^2)/((2^2.3)^6+8^4.3^5)-...

a) A = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)

=> A = \(\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{\left(2^2\right)^6.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{125^3.7^3+5^9.\left(2.7\right)^3}\)

=> A = \(\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{\left(5^3\right)^3.7^3+5^9.2^3.7^3}\)

=> A = \(\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^3\left(1-7\right)}{5^9.7^3+5^9.2^3.7^3}\)

=> A = \(\frac{3-1}{3\left(3+1\right)}-\frac{5^{10}.7^3.\left(-6\right)}{5^9.7^3\left(1+2^3\right)}\)

=> A = \(\frac{2}{3.4}-\frac{5.\left(-6\right)}{9}\)

A = \(\frac{1}{3.2}-\frac{-30}{9}\)

A = \(\frac{1}{6}-\frac{-10}{3}\)

A = \(\frac{1}{6}+\frac{10}{3}=\frac{1}{6}+\frac{20}{6}=\frac{21}{6}\)

=> A = \(\frac{7}{2}=3\frac{1}{2}\)

vậy A = \(3\frac{1}{2}\)

b) ta có:

3ⁿ⁺²-2ⁿ⁺²+3ⁿ-2ⁿ = (3ⁿ⁺²+3ⁿ) - (2ⁿ⁺²-2ⁿ)

= 3ⁿ(9+1) - 2ⁿ(4+1)

= 3ⁿ.10 - 2ⁿ.5

ta thấy: 3ⁿ.10 \(⋮\) 10

2ⁿ là một số chẵn mà 1 số chẵn nhân vs 5 luôn ra kết quả có tận cùng bằng 0 => 2ⁿ.5 \(⋮\) 10

=> 3ⁿ. 10 - 2ⁿ.5 \(⋮\) 10

=> 3ⁿ⁺²-2ⁿ⁺²+3ⁿ-2ⁿ \(⋮\) 10 vs mọi số nguyên dương n ( đpcm)