Ta có :

+) \(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)

+) \(\dfrac{1}{5^2}< \dfrac{1}{4.5}\)

.......................

+) \(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

\(\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+.....+\dfrac{1}{100^2}< \dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{99.100}\)

\(\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+............+\dfrac{1}{100^2}< \dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+......+\dfrac{1}{100}< \dfrac{1}{3}-\dfrac{1}{100}< \dfrac{1}{3}\) \(\left(1\right)\)

Lại có :

+) \(\dfrac{1}{4^2}>\dfrac{1}{4.5}\)

+) \(\dfrac{1}{5^2}>\dfrac{1}{5.6}\)

.....................

+) \(\dfrac{1}{100^2}>\dfrac{1}{100.101}\)

\(\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+........+\dfrac{1}{100^2}>\dfrac{1}{4.5}+\dfrac{1}{5.6}+.........+\dfrac{1}{100.101}\)

\(\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+........+\dfrac{1}{100^2}>\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{100}-\dfrac{1}{101}\)

\(\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+........+\dfrac{1}{100^2}>\dfrac{1}{4}-\dfrac{1}{101}>\dfrac{1}{4}>\dfrac{1}{5}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)

\(\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{100.101}\)

\(\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{100}-\dfrac{1}{101}\)

\(\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4}-\dfrac{1}{101}>\dfrac{1}{4}-\dfrac{1}{20}=\dfrac{1}{5}\)

Lại có \(\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)

\(\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}-\dfrac{1}{100}< \dfrac{1}{3}\)

Vậy \(\dfrac{1}{5}< \dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}\)

\(\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{100\cdot101}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{4}-\dfrac{1}{101}>\dfrac{1}{4}-\dfrac{1}{20}=\dfrac{1}{5}\left(1\right)\)

\(\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{99\cdot100}=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{3}-\dfrac{1}{100}< \dfrac{1}{3}\left(2\right)\) Từ (1) và (2) \(\Rightarrow\dfrac{1}{5}< \dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}< \dfrac{1}{3}\)

T làm biếng lắm; làm C thôi

\(A=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\\ \Rightarrow A< \dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{100}{101}\\ \Rightarrow A^2< \left(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\right).\left(\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{100}{101}\right)\\ =\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{99}{100}.\dfrac{100}{101}\\ =\dfrac{1}{101}< \dfrac{1}{100}\\ \Rightarrow A< \dfrac{1}{10}\)

Làm tương tự ta được A > 1/15

câu a

\(A=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{30}>\dfrac{20}{30}=\dfrac{2}{3}>\dfrac{1}{3}\)

\(A=\left(\dfrac{1}{11}+..+\dfrac{1}{15}\right)+\left(\dfrac{1}{16}+...+\dfrac{1}{30}\right)< 5.\dfrac{1}{10}+25.\dfrac{1}{15}=\dfrac{1}{2}+\dfrac{5}{3}=\dfrac{8}{6}=\dfrac{4}{3}< \dfrac{5}{2}\)

@Ngô Tấn Đạt

Đặt biểu thức trong ngoặc đơn là B

\(5B=1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{98}}+\dfrac{1}{5^{99}}\)

\(\Rightarrow4B=5B-B=1-\dfrac{1}{5^{100}}\Rightarrow B=\dfrac{1}{4}\left(1-\dfrac{1}{5^{100}}\right)\)

\(\Rightarrow A=4.5^{100}.\dfrac{1}{4}\left(\dfrac{5^{100}-1}{5^{100}}\right)+1=\)

\(=5^{100}\)

câu thứ 2 =0 vì (63.1,-21.3,6)=0

MIK muốn hỏi câu đầu tiên

đề đúng ko? ( chỗ 2 cái phân số cuối cùng của vế trái ý)

Đề bài trên sai. Đề đúng: CM: \(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{97}{98}.\dfrac{99}{100}>\dfrac{\sqrt{2}}{20}\).

\(5D=1+\dfrac{1}{5^2}-\dfrac{1}{5^3}+\dfrac{1}{5^4}-\dfrac{1}{5^5}+...+\dfrac{1}{6.5^{99}}\)

\(6D=\dfrac{5^{100}-1}{5^{100}}+\dfrac{1}{6.5^{100}}\)

\(D=\dfrac{\dfrac{5^{100}-1}{5^{100}}+\dfrac{1}{36.5^{100}}}{6}\)