Bài 1: Tìm x, biết:
\(\frac{1}{2.3}x+\frac{1}{3.4}x+\frac{1}{4.5}x+.....+\frac{1}{49.50}x=1\)
Bài 2: Chứng minh rằng:
\(a)A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{100^2}< 2\)
\(b)B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{63}< 6\)
\(c)C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.......\frac{9999}{10000}< \frac{1}{100}\)
Bài 3: Tính tổng:
\(S=\frac{1+2+2^2+2^3+.....+2^{2008}}{1-2^{2009}}\)
Bài 1 :
\(x\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\right)=1\)
\(\Rightarrow x\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)=1\)
\(\Rightarrow x\left(\frac{1}{2}-\frac{1}{50}\right)=1\)
\(\Rightarrow x\cdot\frac{24}{50}=1\)
\(\Rightarrow x=1\div\frac{24}{50}=\frac{25}{12}\)
#Louis
\(\frac{1}{2.3}x+\frac{1}{3.4}x+\frac{1}{4.5}x+...+\frac{1}{49.50}x=1\)
\(\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)x=1\)
\(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)x=1\)
\(\left(\frac{1}{2}-\frac{1}{50}\right)x=1\)
\(\frac{12}{25}x=1\)
Đến đây dễ rồi :)))
Bn tự tính típ nha
Bài 1 :
\(\frac{1}{2.3}x+\frac{1}{3.4}x+\frac{1}{4.5}x+...+\frac{1}{49.50}x=1\)
=> \(x\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)=1\)
=> \(x\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)=1\)
=> \(x\left(\frac{1}{2}-\frac{1}{50}\right)=1\)
=> \(x.\frac{12}{25}=1\)
=> \(x=\frac{25}{12}\)
Study well ! >_<
bài 2: b)B= \(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}\)
=\(1+\left(\frac{1}{2}+\frac{1}{3}\right)+\)\(\left(\frac{1}{4}+...+\frac{1}{7}\right)\)+\(\left(\frac{1}{8}+...+\frac{1}{15}\right)\)+\(\left(\frac{1}{16}+...+\frac{1}{31}\right)\)+\(\left(\frac{1}{32}+...+\frac{1}{63}\right)\)<1+\(\left(\frac{1}{2}+\frac{1}{2}\right)\)+\(\left(\frac{1}{4}+...+\frac{1}{4}\right)\)+\(\left(\frac{1}{8}+...+\frac{1}{8}\right)\)+\(\left(\frac{1}{16}+...+\frac{1}{16}\right)\)+\(\left(\frac{1}{32}+...+\frac{1}{32}\right)\)=\(1+\frac{1}{2}.2+\frac{1}{4}.4\)\(+\frac{1}{8}.8+\frac{1}{16}.16+\frac{1}{32}.32\)=1+1+1+1+1+1=6
vậy B<6
a)
bài 3: đặt A=\(1+2+2^2+2^3+...+2^{2008}\)
=> 2A= \(2+2^2+2^3+...+2^{2009}\)
=> 2A-A=\(2+2^2+2^3+...+2^{2009}\)-\(1-2-2^2-2^3-...-2^{2008}\)
=> A=\(2^{2009}-1\)
thay vào S ta có: S=\(\frac{2^{2009}-1}{1-2^{2009}}\)
vậy S=\(\frac{2^{2009}-1}{1-2^{2009}}\)
\(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9999}{10000}\)
\(\Rightarrow C< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{10000}{10001}\)
\(\Rightarrow C^2< \frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}...\frac{10000}{10001}\)
\(\Rightarrow C^2< \frac{1}{10001}< \frac{1}{10000}=\frac{1}{100^2}\)
Vậy \(C< \frac{1}{100}\left(đpcm\right)\)
