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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)
b: \(\left|x-\dfrac{3}{5}\right|< \dfrac{1}{3}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{3}{5}>-\dfrac{1}{3}\\x-\dfrac{3}{5}< \dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\dfrac{4}{15}< x< \dfrac{14}{15}\)
c: \(\left|x+\dfrac{11}{2}\right|>-5.5\)
mà \(\left|x+\dfrac{11}{2}\right|\ge0\forall x\)
nên \(x\in R\)
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+........+\dfrac{1}{100^2}\)
Ta có :
\(\dfrac{1}{5^2}< \dfrac{1}{4.5}\)
\(\dfrac{1}{6^2}< \dfrac{1}{5.6}\)
...................
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Leftrightarrow\dfrac{1}{5^2}+\dfrac{1}{6^2}+....+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+.......+\dfrac{1}{99.100}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+......+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{4}-\dfrac{1}{100}=\dfrac{6}{25}\)
Mà \(\dfrac{1}{6}< \dfrac{5}{26}< \dfrac{1}{4}\)
Mà \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+.........+\dfrac{1}{100^2}< \dfrac{6}{25}\)
\(\Leftrightarrow\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+.......+\dfrac{1}{100^2}< \dfrac{1}{4}\left(đpcm\right)\) \(\left(1\right)\)
a: 2x(x-1/7)=0
=>x(x-1/7)=0
=>x=0 hoặc x=1/7
b: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)
nên \(x=\dfrac{-1}{4}:\dfrac{7}{20}=\dfrac{-20}{4\cdot7}=\dfrac{-5}{7}\)
c: \(\Leftrightarrow\dfrac{41}{9}:\dfrac{41}{18}-7< x< \left(3.2:3.2+\dfrac{45}{10}\cdot\dfrac{31}{45}\right):\left(-21.5\right)\)
\(\Leftrightarrow2-7< x< \dfrac{\left(1+3.1\right)}{-21.5}\)
\(\Leftrightarrow-5< x< \dfrac{-41}{215}\)
mà x là số nguyên
nên \(x\in\left\{-4;-3;-2;-1\right\}\)
a) \(\dfrac{x}{48}=-\dfrac{4}{7}\Rightarrow x=-\dfrac{192}{7}\)
b) \(\left(x+\dfrac{4}{5}\right)-\dfrac{2}{5}=\dfrac{3}{5}\Rightarrow x+\dfrac{4}{5}=1\)
\(\Rightarrow x=\dfrac{1}{5}\)
c) \(2\left|x-1\right|^2=72\Rightarrow\left|x-1\right|^2=36\)
\(\Rightarrow\left|x-1\right|=6\)
TH1: x - 1 = -6 => x = -5
TH2: x - 1 = 6 => x = 7
e) \(\dfrac{x}{2,5}=\dfrac{4}{5}\Rightarrow x=2\)
f) | x - 2 | = 1 + 4 = 5
TH1: x - 2 = -5 => x = -3
TH2: x - 2 = 5 => x = 7
a) \(\dfrac{x}{48}=\dfrac{-4}{7}\)
⇒ x.7=48.(-4)
7x = -192
x=\(\dfrac{-192}{7}\) Vậy x=\(\dfrac{-192}{7}\)
b) \(\left(x+\dfrac{4}{5}\right)-\dfrac{2}{5}=\dfrac{3}{5}\)
\(\left(x+\dfrac{4}{5}\right)=\dfrac{3}{5}+\dfrac{2}{5}\)
\(x+\dfrac{4}{5}=1\)
\(x=1-\dfrac{4}{5}\)
\(x=\dfrac{1}{5}\)
c) chưa từng gặp dạng với giá trị tuyệt đối sory
d) \(\dfrac{1}{6}x-\dfrac{2}{3}=2\)
\(\dfrac{1}{6}x=2+\dfrac{2}{3}\)
\(\dfrac{1}{6}x=\dfrac{8}{3}\)
\(x=\dfrac{8}{3}:\dfrac{1}{6}\)
\(x=16\)
e) \(\dfrac{x}{2,5}=\dfrac{4}{5}\)
=> x.5 = 4.2,5
5x=10
x=10:5
x=2
f) |x-2|-4=1
|x-2|=1+4
|x-2|=5
=>\(\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=5+2\\x=-5+2\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
đôi khi cũng có sai sót , hãy xem lại thật kĩ
\(C=\dfrac{5}{5\cdot8\cdot11}+\dfrac{5}{8\cdot11\cdot14}+...+\dfrac{5}{302\cdot305\cdot308}\\ =\dfrac{5}{6}\cdot\left(\dfrac{6}{5\cdot8\cdot11}+\dfrac{6}{8\cdot11\cdot14}+...+\dfrac{6}{302\cdot305\cdot308}\right)\\ =\dfrac{5}{6}\cdot\left(\dfrac{1}{5\cdot8}-\dfrac{1}{8\cdot11}+\dfrac{1}{8\cdot11}-\dfrac{1}{11\cdot14}+...+\dfrac{1}{302\cdot305}-\dfrac{1}{305\cdot308}\right)\\ =\dfrac{5}{6}\cdot\left(\dfrac{1}{40}-\dfrac{1}{305\cdot308}\right)\\ =\dfrac{5}{6}\cdot\dfrac{1}{40}-\dfrac{5}{6}\cdot\dfrac{1}{305\cdot308}\\ =\dfrac{1}{48}-\dfrac{5}{6\cdot305\cdot308}\\ \dfrac{5}{6\cdot305\cdot308}>0\Rightarrow\dfrac{1}{48}-\dfrac{5}{6\cdot305\cdot308}< \dfrac{1}{48}\)