Question

Bài 1

a) thực hiện phép tính A=\(\frac{2^{12}\times3^5-4^6\times9^2}{\left(2^2\times3\right)^6+8^4\times3^5}-\frac{5^{10}\times7^3+25^5\times49^2}{\left(125\times7\right)^3+5^9\times14^3}\)

b) CMR: Với mọi số nguyên dương n thì:\(3^{n+2}-2^{n+2}+3^n-2^n\) chia hết cho 10

Answer 1

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

\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3+5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)

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

\(=\frac{2}{12}-\frac{5.8}{9}=\frac{1}{6}-\frac{40}{9}=\frac{-77}{18}\)

b ) 3ⁿ⁺² - 2ⁿ⁺² + 3ⁿ - 2ⁿ

= ( 3ⁿ⁺² + 3ⁿ ) - ( 2ⁿ⁺² + 2ⁿ )

= 3ⁿ ( 3² + 1 ) - 2ⁿ ( 2² + 1 )

= 3ⁿ.10 - 2^n-1.2.5

= 3ⁿ.10 - 2^n-1.10

= ( 3ⁿ - 2^n-1 ).10 chia hết cho 10 ( đpcm )