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\(=\frac{2-1}{2!}+\frac{3-1}{3!}+...+\frac{n-1}{n!}\)
\(=\left(1-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+...+\left(\frac{1}{n-1!}-\frac{1}{n!}\right)\)
\(=1-\frac{1}{n!}< 1\)
\(\Rightarrow\frac{1}{2!}+\frac{1}{3!}+...+\frac{n-1}{n!}< 1\)
bài 3.\(\frac{x-7}{3}=\frac{4x-1}{2}\Leftrightarrow2x-14=12x-3\\ \Leftrightarrow10x=-11\\ \Leftrightarrow x=\frac{-11}{10}\)
Bài 1 :
ta thấy :
\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};......;\frac{1}{\left(n-1\right)^2}< \frac{1}{\left(n-2\right).\left(n-1\right)};\frac{1}{n^2}< \frac{1}{\left(n-1\right).n}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{\left(n-1\right)^2}+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{\left(n-2\right).\left(n-1\right)}+\frac{1}{\left(n-1\right).n}\)
mà :
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{\left(n-2\right).\left(n-1\right)}+\frac{1}{\left(n-1\right).n}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{n-1}-\frac{1}{n}\)
=\(1-\frac{1}{n}\)<1
=>A<1
Bài 3 :
\(\frac{x-7}{3}=\frac{4x-1}{2}\)
=>\(\frac{2.\left(x-7\right)}{6}=\frac{3\left(4x-1\right)}{6}\)
=>\(2\left(x-7\right)=3\left(4x-1\right)\)
=>\(x-7=\frac{3}{2}.\left(4x-1\right)\)
=>\(\frac{x-7}{4x-1}=\frac{3}{2}\)
\(=>\left\{{}\begin{matrix}x-7=3=>x=10\\4x-1=2=>4x=3=>x=\frac{3}{4}\end{matrix}\right.\)
Vậy x ∈{\(10;\frac{3}{4}\)}
\(N=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)
\(N=\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)< \frac{1}{2^2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{...1}{\left(n-1\right).n}\right)\)
\(N< \frac{1}{4}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)
\(N< \frac{1}{4}.\left(1-\frac{1}{n}\right)< \frac{1}{4}.1=\frac{1}{4}\)
=> \(N< \frac{1}{4}\)(đpcm)
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