\(=\dfrac{5^{102}.\left(3^2\right)^{1009}}{3^{2018}.\left(5^2\right)^{50}}\)
\(=\dfrac{5^{102}.3^{2018}}{3^{2018}.5^{100}}=5^2=25\)
Đúng 0
Bình luận (2)
Violympic toán 7
Các câu hỏi tương tự
Thực hiện phép tính
\(\dfrac{5^{102}.9^{1009}}{3^{2018}.25^{50}}\)
\(\dfrac{\sqrt{\dfrac{9}{4}-3^{-1}+2018^0}}{25\%+1\dfrac{1}{4}-1,3}-\dfrac{\left(\dfrac{-1}{2}\right)^2-\sqrt{\dfrac{4}{9}}+0,4}{0,6-\dfrac{2}{3}.\left(\dfrac{-1}{4}-\dfrac{1}{2}\right)}\)
1) \(25^{10}.\left(\dfrac{1}{5}\right)^{20}+\left(\dfrac{-3}{4}\right)^8.\left(\dfrac{-4}{3}\right)^8-2018^0\)
2) \(\left(\dfrac{5}{2}-\dfrac{4}{3}\right).\dfrac{6}{7}+\left(\dfrac{-3}{2}\right)^5:\left(\dfrac{-3}{2}\right)^3\)
3) \(\dfrac{4^5.9^4-2.6^9}{3^8.2^{10}+6^8.20}\)
1) Cho 2 số hữu tỉ x, y có tổng bằng 4. Chứng minh rằng x.y ≤ 4
2) Cho 3 số hữu tỉ dương a, b, c thỏa mãn: \(\dfrac{a+b-c}{a}=\dfrac{b+c-a}{b}=\dfrac{c+a-b}{c}\)
Tính giá trị của biểu thức P = \(\dfrac{a^{1008}.b^{1009}.c}{a^{2018}+b^{2018}+c^{2018}}\)
\(\left(\dfrac{1}{3}\right)^{50}.\left(-9\right)^{25}-\dfrac{2}{3}:4\)
\(\dfrac{\sqrt{\dfrac{9}{4}}-3^{-1}+2018^0}{25\%+1\dfrac{1}{4}-1,3}-\dfrac{\left(\dfrac{-1}{2}\right)^2-\sqrt{\dfrac{4}{9}+0,4}}{0,6-\dfrac{2}{3}\left(\dfrac{-1}{4}-\dfrac{1}{2}\right)}\)
Nhớ giải chi tiết nha 
\(\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^3\cdot3\right)+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\) .Tính tổng
Cho S=\(\dfrac{1}{5^2}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+\dfrac{4}{5^4}+...+\dfrac{2017}{5^{2017}}+\dfrac{2018}{5^{2018}}\).Chứng minh S<\(\dfrac{1}{3}\)
1) \(\left(\dfrac{1}{3}\right)^{50}.90^{25}-\dfrac{2}{3}:4\)
2) \(10.\sqrt{0,01}.\sqrt{\dfrac{16}{9}}+\sqrt{49}-\dfrac{1}{6}.\sqrt{4}\)