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\(=\frac{1}{4}.\left(\frac{17.4}{1.3.5}+\frac{17.4}{3.5.7}+\frac{17.4}{5.7.9}+...+\frac{17.4}{47.49.51}\right)\)
\(=\frac{17}{4}\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{47.49}-\frac{1}{49.51}\right)\)
\(=\frac{17}{4}\left(\frac{1}{3}-\frac{1}{2499}\right)=\frac{17}{4}.\frac{832}{2499}=\frac{208}{147}\)
Bài làm:
Ta có: \(A=\frac{1}{1.3.5}+\frac{1}{3.5.7}+...+\frac{1}{47.49.51}\)
\(A=\frac{1}{4}\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+...+\frac{4}{47.49.51}\right)\)
\(A=\frac{1}{4}\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{47.49}-\frac{1}{49.51}\right)\)
\(A=\frac{1}{4}\left(\frac{1}{3}-\frac{1}{49.51}\right)\)
\(A=\frac{1}{12}-\frac{1}{4.49.51}< \frac{1}{12}\)
Vậy \(A< \frac{1}{12}\)
Từ đề bài suy ra\(4A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+...+\frac{4}{47.49.51}\)
\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{47.49}-\frac{1}{49.51}=\frac{1}{3}-\frac{1}{49.51}< \frac{1}{3}\)
\(\Rightarrow A< \frac{1}{12}\left(đpcm\right)\)
a, Ta có : x-13\(\ne0\)
Mà để phân số A có giá trị nhỏ nhất => A=1
=> 17/x-13=1
=> x-13=17
=> x=30
:)
đúng sai làm thì bieets
a: \(A=\left(\dfrac{15}{34}+\dfrac{9}{34}-1-\dfrac{15}{17}\right)+\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)
\(=\left(\dfrac{12}{17}-1-\dfrac{15}{17}\right)+1\)
\(=\dfrac{-20}{17}+1=\dfrac{-3}{17}\)
b: \(B=\dfrac{-5}{3}\cdot16\dfrac{2}{7}-\dfrac{-5}{3}\cdot28\dfrac{2}{7}\)
\(=\dfrac{-5}{3}\left(16+\dfrac{2}{7}-28-\dfrac{2}{7}\right)=\dfrac{-5}{3}\cdot\left(-12\right)=20\)
c: \(C=25\cdot\dfrac{-1}{27}+\dfrac{1}{5}-2\cdot\dfrac{1}{4}-\dfrac{1}{2}\)
\(=\dfrac{-25}{27}+\dfrac{1}{5}-1\)
\(=\dfrac{-125+27-135}{135}=\dfrac{-233}{135}\)
= 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + .....+1/98.99 - 1/99.100
= 1/2 - 1/9900
= 4949/9900
k cho minh nha
chuc ban hoc tot
\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\)
\(=2.\frac{1}{1.2.3}+2.\frac{1}{2.3.4}+...+2.\frac{1}{98.99.100}\)
\(=2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\right)\)
\(=2.\left[\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{98.99}-\frac{1}{99.100}\right)\right]\)
\(=2.\left[\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\right]\)
\(=2.\left[\frac{1}{2}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\right]\)
\(=2.\left[\frac{1}{2}.\left(1-\frac{1}{100}\right)\right]\)
\(=2.\left(\frac{1}{2}.\frac{99}{100}\right)\)
\(=\left(2.\frac{1}{2}\right).\frac{99}{100}\)
\(=1.\frac{99}{100}\)
\(=\frac{99}{100}\)
a) \(A=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{98\cdot99\cdot100}\)
\(A=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\)
\(A=\frac{1}{2}-\frac{1}{99\cdot100}=\frac{1}{2}-\frac{1}{9900}=\frac{4949}{9900}\)
b) \(B=\frac{17}{1\cdot3\cdot5}+\frac{17}{3\cdot5\cdot7}+\frac{17}{5\cdot7\cdot9}+...+\frac{17}{47\cdot49\cdot51}\)
\(B=\frac{17}{4}\left(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+...+\frac{4}{47\cdot49\cdot51}\right)\)
\(B=\frac{17}{4}\left(\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{47\cdot49}-\frac{1}{49\cdot51}\right)\)
\(B=\frac{17}{4}\left(\frac{1}{3}-\frac{1}{2499}\right)=\frac{17}{4}\cdot\frac{832}{2499}=\frac{208}{147}\)