Tính nhanh :
\(248.\left(\left(\frac{1}{1.1985}+\frac{1}{2.1984}+....+\frac{1}{16.2000}\right):\left(\frac{1}{1.17}+\frac{1}{2.18}+...+\frac{1}{1984.2000}\right)\right)\)
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So Sánh
\(A=127.\left(\frac{1}{1.1985}+\frac{1}{2.1986}+...+\frac{1}{16.2000}\right)\)
\(B=\frac{1}{1.17}+\frac{1}{2.18}+\frac{1}{3.19}+...+\frac{1}{1984.2000}\)
So sánh
A=124.\(\left(\frac{1}{1.1985}+\frac{1}{2.1986}+\frac{1}{3.1987}+...+\frac{1}{16.2000}\right)\)
Và B=\(\frac{1}{1.17}+\frac{1}{2.18}+\frac{1}{3.19}+...+\frac{1}{1984.2000}\)
So sánh
E=\(124.\left(\frac{1}{1.1985}+\frac{1}{2.1986}+\frac{1}{3.1987}+...+\frac{1}{16.2000}\right)\)
Và F=\(\frac{1}{1.17}+\frac{1}{2.18}+\frac{1}{3.19}+...+\frac{1}{1984.2000}\)
So sánh
E=124.\(\left(\frac{1}{1.1985}+\frac{1}{2.1986}+\frac{1}{3.1987}+...+\frac{1}{16.2000}\right)\)
Và F=\(\frac{1}{1.17}+\frac{1}{2.18}+\frac{1}{3.19}+...+\frac{1}{1984.2000}\)
Câu hỏi của Trương Nguyễn Bảo Trân - Toán lớp 6 - Học toán với OnlineMath tham khảo
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so sánh : A = \(124.\left(\frac{1}{1.1985}+\frac{1}{2.1986}+\frac{1}{3.1987}+...+\frac{1}{16.2000}\right)\)
B=\(\frac{1}{1.17}+\frac{1}{2.18}+\frac{1}{3.19}+...+\frac{1}{1984.2000}\)
\(A=124\left(\frac{1}{1.1985}+\frac{1}{2.1986}+\frac{1}{3.1987}+...+\frac{1}{16.2000}\right)\)
\(=\frac{124}{1984}.\left(1-\frac{1}{1985}+\frac{1}{2}-\frac{1}{1986}+...+\frac{1}{16}-\frac{1}{2000}\right)\)
\(=\frac{1}{16}\left[\left(1+\frac{1}{2}+...+\frac{1}{16}\right)-\left(\frac{1}{1985}+\frac{1}{1986}+...+\frac{1}{2000}\right)\right]\)
Và \(B=\frac{1}{1.17}+\frac{1}{2.18}+...+\frac{1}{1984.2000}\)
\(=\frac{1}{16}\left[\left(1-\frac{1}{17}+\frac{1}{2}-\frac{1}{18}+...+\frac{1}{1984}-\frac{1}{2000}\right)\right]\)
\(=\frac{1}{16}\left[\left(1+\frac{1}{2}+...+\frac{1}{1984}\right)-\left(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{2000}\right)\right]\)
= \(\frac{1}{16}\) . \(\left[\left(1+...+\frac{1}{16}\right)+\left(\frac{1}{17}+...+\frac{1}{1984}-\frac{1}{17}-...-\frac{1}{1984}\right)-\left(\frac{1}{1985}+...+\frac{1}{2000}\right)\right]\)
= \(=\frac{1}{16}\left[\left(1+\frac{1}{2}+...+\frac{1}{16}\right)-\left(\frac{1}{1985}+\frac{1}{1986}+...+\frac{1}{2000}\right)\right]\)
Vậy A = B
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So sánh 2 biểu thức:
\(A=124\left(\frac{1}{1.1985}+\frac{1}{2.1986}+\frac{1}{3.1987}+...+\frac{1}{16.2000}\right)\)
\(B=\frac{1}{1.17}+\frac{1}{2.18}+\frac{1}{3.19}+...+\frac{1}{1984.2000}\)
So sánh 2 biểu thức:
A = 124.\(\left(\frac{1}{1.1985}+\frac{1}{2.1986}+\frac{1}{3.1987}+.....+\frac{1}{16.2000}\right)\)
B = \(\frac{1}{1.17}+\frac{1}{2.18}+\frac{1}{3.19}+......+\frac{1}{1984.2000}\)
So sánh 2 biểu thức:
\(A=124.\left(\frac{1}{1.1985}+\frac{1}{2.1986}+\frac{1}{3.1987}+...+\frac{1}{16.2000}\right)\)
\(B=\frac{1}{1.17}+\frac{1}{2.18}+\frac{1}{3.19}+...+\frac{1}{1984.2000}\)
So sánh A và B biết:
\(a=124.\left(\frac{1}{1.1985}+\frac{1}{2.1986}+\frac{1}{3.1987}+...+\frac{1}{16.2000}\right)\)
\(b=\frac{1}{1.17}+\frac{1}{2.18}+\frac{1}{3.19}+...+\frac{1}{1984.2000}\)
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