Câu 1: Chứng minh rằng:
\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+\(\frac{1}{5^2}\)+\(\frac{1}{6^2}\).....+\(\frac{1}{100^2}\)<\(\frac{1}{2}\)
Câu 2: Rút gọn biểu thức:
A=\(\left(\frac{1}{2}+1\right)\).\(\left(\frac{1}{3}+1\right)\).\(\left(\frac{1}{4}+ 1\right)\)....\(\left(\frac{1}{98}+1\right)\).\(\left(\frac{1}{99}+1\right)\)
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1. Giải:
Ta có:
\(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)<\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)=\(\frac{1}{2}-\frac{1}{100}\)<\(\frac{1}{2}\) => \(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)<\(\frac{1}{2}\)
2. Giải
A = \(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)
A =
Câu 1:
Đặt: \(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+....+\frac{1}{100^2}\)
\(=\frac{1}{3.3}+\frac{1}{4.4}+\frac{1}{5.5}+\frac{1}{6.6}+....+\frac{1}{100.100}\)
\(A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+.....+\frac{1}{99.100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{100}\)
\(\Rightarrow A< \frac{49}{100}< \frac{50}{100}=\frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2}\)
Vậy:.............
Câu 2:
\(\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)...\left(\frac{1}{98}+1\right)\left(\frac{1}{99}+1\right)\)
\(=\left(\frac{1}{2}+\frac{2}{2}\right)\left(\frac{1}{3}+\frac{3}{3}\right)\left(\frac{1}{4}+\frac{4}{4}\right)...\left(\frac{1}{98}+\frac{98}{98}\right)\left(\frac{1}{99}+\frac{99}{99}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{99}{98}.\frac{100}{99}\)
\(=\frac{3.4.5....99.100}{2.3.4...98.99}\)
\(=\frac{100}{2}=50\)
