- Tim x
\(\left[\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+..........+\frac{1}{98.99.100}\right].x=\frac{49}{200}\)
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=(1/1.2-1/2.3+1/2.3-1/3.4+..........+1/98.99-1/99.100).X =-3
=>(1/1.2-1/99.100).x=-3
<=>98/99.x=-3
=>x=98/99 : 3
=98/297
ko biet dung ko
hihi
\(A=2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\right)\)
\(A=2.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(A=2.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(A=2\cdot\frac{4949}{9900}=\frac{4949}{4950}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{98.99.100}=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{k}=\frac{1}{2}\Rightarrow k=2\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(=\frac{1}{2}.\frac{4949}{9900}\)
\(=\frac{4949}{19800}\)
Ta có
Z = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ... + 1/98.99.100
2Z = 2/1.2.3 + 2/2.3.4 + 2/3.4.5 + ... + 2/98.99.100
2Z = 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/4.5 + ... + 1/98.99 - 1/99.100
2Z = 1/1.2 - 1/99.100
2Z = 4949/9900
=> Z = 4949/19800
=> 4949/19800 . x = 49/200
x = 49/200 : 4949/19800
x = 99/101
Vậy x = 99/101
Ủng hộ nha