1. Tính nhanh:
\(A=\dfrac{\left(12.8\right)^3.5}{2^9.15^2}\)
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CM
25 tháng 9 2017
A=\(\dfrac{12^3\cdot8^3\cdot5}{2^9\cdot\left(3\cdot5\right)^2}\)
A=\(\dfrac{\left(2^2\cdot3\right)^3\cdot\left(2^3\right)^3\cdot5}{2^9\cdot3^2\cdot5^2}\)
A=\(\dfrac{\left(2^2\right)^3\cdot3^3\cdot2^9\cdot5}{2^9\cdot3^2\cdot5^2}\)
A=\(\dfrac{2^6\cdot3^3\cdot2^9\cdot5}{2^9\cdot3^2\cdot5^2}\)
A=\(\dfrac{2^6\cdot3}{5}\)
A=\(\dfrac{64\cdot3}{5}\)
A=\(\dfrac{192}{5}\)
Vậy A=\(\dfrac{192}{5}\)
Tích cho mình nhé!
9 tháng 3 2021
\(A=\dfrac{1}{2}\left(2.\dfrac{2}{3}\right)\left(\dfrac{3}{2}.\dfrac{3}{4}\right)\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)
\(=\dfrac{2016}{2017}\)
25 tháng 5 2022
\(A=\dfrac{1}{2}.\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)....\left(\dfrac{1}{2015.2017}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right).....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)
\(=\dfrac{2016}{2017}\)
MH
17 tháng 10 2023
\(A=\dfrac{1}{2}\left(\dfrac{2.2}{1.3}\right).\left(\dfrac{3.3}{2.4}\right)...\left(\dfrac{2020.2020}{2019.2021}\right)\)
\(=\dfrac{1.2.2.3.3...2020.2020}{1.2.2.3.3.4.4...2019.2021}\)
\(=\dfrac{1}{2021}\)
17 tháng 10 2023
\(A=\dfrac{1}{2}\cdot\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\left(1+\dfrac{1}{3\cdot5}\right)...\left(1+\dfrac{1}{2019\cdot2021}\right)\)
\(A=\dfrac{1}{2}\left(1+\dfrac{1}{2^2-1}\right)\left(1+\dfrac{1}{3^2-1}\right)\left(1+\dfrac{1}{4^2-1}\right)...\left(1+\dfrac{1}{2020^2-1}\right)\)
\(A=\dfrac{1}{2}\cdot\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\cdot\left(3+1\right)}...\left(\dfrac{2020^2}{\left(2020-1\right)\cdot\left(2020+1\right)}\right)\)
\(A=\dfrac{1}{2}\cdot\dfrac{2}{1}\cdot\dfrac{2}{3}\cdot\dfrac{3}{2}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2020}{2021}\)
\(A=\dfrac{1}{2}\cdot\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2021}\)
\(A=\dfrac{1}{2}\cdot2020\cdot\dfrac{2}{2021}=\dfrac{2020}{2021}\)
26 tháng 4 2017
a. \(\dfrac{2^3.5^2.7^2.3^7}{49.5^3.3^6.11}\)
= \(\dfrac{2^3.3}{5.11}=\dfrac{24}{55}\)
b. \(4.\left(\dfrac{-1}{2}\right)^3-2.\left(\dfrac{-1}{2}\right)^2+3\left(\dfrac{-1}{2}\right)+1\)
=\(-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{2}\)
= 3\(\left(\dfrac{-1}{2}\right)\)
=\(\dfrac{-3}{2}\)
12 tháng 7 2023
\(=\dfrac{1}{2}\cdot\dfrac{2^2-1+1}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2-1+1}{\left(3-1\right)\left(3+1\right)}\cdot...\cdot\dfrac{2016^2-1+1}{\left(2016-1\right)\left(2016+1\right)}\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot...\cdot\dfrac{2016}{2015}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2016}{2017}\)
\(=\dfrac{1}{2}\cdot2016\cdot\dfrac{2}{2017}=\dfrac{2016}{2017}\)
6 tháng 4 2017
a) \(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{780}\right)\)
\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}.....\dfrac{779}{780}\)\(=\)
25 tháng 1 2021
Thuật toán:
Bước 1: Nhập n
Bước 2: i←1; a←0;
Bước 3: a←a+1/(i*(i+2));
Bước 4: i←i+1;
Bước 5: Nếu i<=n thì quay lại bước 3
Bước 6: xuất a
Bước 7: Kết thúc
Viết chương trình:
uses crt;
var a:real;
i,n:longint;
begin
clrscr;
write('Nhap n='); readln(n);
a:=0;
for i:=1 to n do
a:=a+1/(i*(i+2));
writeln(a:4:2);
readln;
end.
25 tháng 8 2023
\(R=\dfrac{\sqrt{\left(-\dfrac{2}{5}\cdot\dfrac{-5}{8}\right)^3\cdot5^2}}{\sqrt[3]{\dfrac{-3^3}{4^3}\cdot\dfrac{5^2}{2^6\cdot3^2}\cdot\dfrac{5^4}{3^4}}}\)
\(=\dfrac{\sqrt{\left(\dfrac{1}{4}\right)^3\cdot5^2}}{\sqrt[3]{\dfrac{-1}{3^3}\cdot\dfrac{25^3}{16^3}}}=\dfrac{5}{8}:\dfrac{-5}{3\cdot4}=\dfrac{5}{8}\cdot\dfrac{3\cdot4}{-5}=-\dfrac{3}{2}\)
\(A=\dfrac{\left(12.8\right)^3.5}{2^9.15^2}=\dfrac{\left(2^2.3.2^3\right)^3.5}{2^9.\left(3.5\right)^2}=\dfrac{2^6.3^3.2^9.5}{2^9.3^2.5^2}=\dfrac{2^{15}.3^3.5}{2^9.3^2.5^2}=\dfrac{2^6.3.1}{1.1.5}=\dfrac{192}{5}\)