S=\(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}....\dfrac{199}{200}\)
CMR S2<\(\dfrac{1}{200}\)
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Ta có:\(C=\dfrac{1}{2}.\dfrac{3}{4}.....\dfrac{199}{200}\)
\(\Rightarrow C< \dfrac{2}{3}.\dfrac{4}{5}.....\dfrac{200}{201}\)
\(\Rightarrow C^2< \dfrac{2}{3}.\dfrac{4}{5}.....\dfrac{200}{201}.\dfrac{1}{2}.\dfrac{3}{4}.....\dfrac{199}{200}\)
\(\Rightarrow C^2< \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.....\dfrac{199}{200}.\dfrac{200}{201}\)
\(\Rightarrow C^2< \dfrac{1}{201}\) (đpcm)
Ta có: \(\dfrac{1}{101}>\dfrac{1}{200}\)
Tương tự ta có: \(\dfrac{1}{102}>\dfrac{1}{200}\) ;....; \(\dfrac{1}{199}>\dfrac{1}{200}\)
\(\Rightarrow\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{199}+\dfrac{1}{200}>\dfrac{1}{200}.100\)
\(\Leftrightarrow\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{199}+\dfrac{1}{200}>\dfrac{100}{200}\)
\(\Leftrightarrow\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{199}+\dfrac{1}{200}>\dfrac{1}{2}\left(đpcm\right)\)
\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{199}-\dfrac{1}{200}\)
\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{199}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+..+\dfrac{1}{200}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{199}+\dfrac{1}{200}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{200}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{199}+\dfrac{1}{200}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)\)
\(=\dfrac{1}{101}+...+\dfrac{1}{199}+\dfrac{1}{200}\)
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!
Bài 1:
a: Sửa đề: 1/3^200
1/2^300=(1/8)^100
1/3^200=(1/9)^100
mà 1/8>1/9
nên 1/2^300>1/3^200
b: 1/5^199>1/5^200=1/25^100
1/3^300=1/27^100
mà 25^100<27^100
nên 1/5^199>1/3^300
a, tổng các tử và mẫu mỗi phân sô trên đều bằng 200
b, \(A=\dfrac{1}{199}+\dfrac{2}{198}+\dfrac{3}{197}+...+\dfrac{198}{2}+\dfrac{199}{1}\)
\(A=\dfrac{200}{199}+\dfrac{200}{198}+...+\dfrac{200}{2}+\dfrac{200}{200}\)
\(A=200\left(\dfrac{1}{199}+\dfrac{1}{198}+...+\dfrac{1}{2}+\dfrac{1}{200}\right)\)(đpcm)
\(S^2=\left(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{199}{200}\right)\left(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{199}{200}\right)\\ \text{Ta có:}\\ \dfrac{1}{2}< \dfrac{2}{3}\\ \dfrac{3}{4}< \dfrac{4}{5}\\ \dfrac{5}{6}< \dfrac{6}{7}\\ ...\\ \dfrac{199}{200}< \dfrac{200}{201}\\ \Rightarrow S^2< \left(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{199}{200}\right)\left(\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}\cdot...\cdot\dfrac{200}{201}\right)\\ \Leftrightarrow S^2< \dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{199}{200}\cdot\dfrac{200}{201}\\ \Leftrightarrow S^2< \dfrac{1\cdot2\cdot3\cdot...\cdot200}{2\cdot3\cdot4\cdot...\cdot201}\\ \Leftrightarrow S^2< \dfrac{1}{201}< \dfrac{1}{200}\)
Vậy ...