\(j,\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{53.55}=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{53}-\dfrac{1}{55}=\dfrac{1}{5}-\dfrac{1}{55}=\dfrac{11}{55}-\dfrac{1}{55}=\dfrac{10}{55}=\dfrac{2}{11}\\ k,\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{99}{100}=\dfrac{1}{100}.\dfrac{2}{2}.\dfrac{3}{3}...\dfrac{99}{99}=\dfrac{1}{100}.1.1...1=\dfrac{1}{100}\)

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

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

Lời giải:

Gọi phân số vế trái là $A$. Gọi tử số là $T$. Xét mẫu số:
\(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)

\(=1-\frac{1}{2}+1-\frac{1}{3}+1-\frac{1}{4}+....+1-\frac{1}{100}\)

\(=99-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=100-(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100})\)

\(=\frac{1}{2}\left[200-(3+\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100})\right]=\frac{1}{2}T\)

$\Rightarrow A=\frac{T}{\frac{1}{2}T}=2$ 

Ta có đpcm.

Giải:

Vì \(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}=2\) nên phần tử gấp 2 lần phần mẫu

Ta có:

\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)  

\(=\dfrac{2.\left[100-\left(\dfrac{3}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{100}\right)\right]}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)

\(=\dfrac{2.\left[\left(2-\dfrac{3}{2}\right)+\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{1}{5}\right)+...+\left(1-\dfrac{1}{100}\right)\right]}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\) 

\(=\dfrac{2.\left(\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{4}{5}+...+\dfrac{99}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\) 

\(=2\) 

Vậy \(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}=2\left(đpcm\right)\) 

Chúc bạn học tốt!

Ta có: \(M=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(1+\dfrac{2}{98}\right)+\left(1+\dfrac{3}{97}\right)+\left(1+\dfrac{4}{96}\right)+...+\left(1+\dfrac{98}{2}\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{98}+\dfrac{100}{97}+...+\dfrac{100}{1}+\dfrac{100}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

=100

Ta có: \(N=\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)

\(=\dfrac{\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{90}{98}\right)+\left(1-\dfrac{91}{99}\right)+\left(1-\dfrac{92}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{99}+\dfrac{8}{100}}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{8}{\dfrac{1}{5}}=40\)

\(\Leftrightarrow\dfrac{M}{N}=\dfrac{100}{40}=\dfrac{5}{2}\)

câu thứ 2 =0 vì (63.1,-21.3,6)=0

MIK muốn hỏi câu đầu tiên

a)

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}\\ < \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{29}-\dfrac{1}{30}\\ =1-\dfrac{1}{30}=\dfrac{29}{30}< 1\left(dpcm\right)\)

b)

 \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}=\dfrac{1}{10}+\left(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\\ >\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{1}{10}+\dfrac{90}{100}\\ =\dfrac{110}{100}>1\left(đpcm\right).\)

c)

\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}\\ =\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}\right)+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{17}\right)\\ < \dfrac{1}{5}.5+\dfrac{1}{8}.8=1+1=2\left(đpcm\right)\)

d) tương tự câu 1

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