Question

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

b) \(\frac{1}{1.6}\)+\(\frac{1}{6.11}\)+\(\frac{1}{11.6}\)+...+\(\frac{1}{96.101}\)

Answer 1

a)

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

\(=\frac{2^{12}\left(3^5-3^4\right)}{2^{12}\left(3^6+3^5\right)}-\frac{5^{10}\left(7^3-7^4\right)}{5^9.7^3\left(1+2^3\right)}\)

\(=\frac{3^5-3^4}{3^6+3^5}-\frac{5\left(7^3-7^4\right)}{7^3.3^2}\)

=\(\frac{3^4\left(3-1\right)}{^{ }3^4\left(9+3\right)}-\frac{5.7^3-5.7^4}{7^3.3^2}\)

=\(\frac{1}{6}-\frac{7^3.5\left(1-7\right)}{7^3.3^2}=\frac{1}{6}-\frac{30}{9}=-\frac{19}{6}\)

Vậy A=\(-\frac{19}{6}\)

Answer 2

câu b lúc nã mk làm sai rui

dây mới đúng

=\(\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{96}-\frac{1}{101}\right)\)

=\(\frac{1}{5}\left(1-\frac{1}{101}\right)=\frac{1}{5}.\frac{100}{101}=\frac{20}{101}\)

Answer 3

bạn chỉ mk cách viết phân số rồi mk giải cho nhé

Answer 4

b) \(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+....+\frac{1}{96.101}\)

=\(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+....+\frac{1}{96}-\frac{1}{101}\)

=\(1-\frac{1}{101}=\frac{100}{101}\)

Answer 5

bài b ở phân số thứ 3 mẫu phải là 11.16