Chứng minh rằng:
\(\frac{3}{1^2\times2^2}+\frac{5}{2^2\times3^2}+\frac{7}{3^2\times4^2}+......+\frac{19}{9^2\times10^2}\) \(<1\)
\(C=\frac{3}{1^2\times2^2}+\frac{5}{2^2\times3^2}+\frac{7}{3^2\times4^2}+........+\frac{19}{9^2\times10^2}\)
So sánh C và D với D=1
=> \(C=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)
C = \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+....+\frac{1}{81}-\frac{1}{100}\)
C = \(1-\frac{1}{100}
\(A=\frac{3}{1\times1\times2\times2}+\frac{5}{2\times2\times3\times3}+...+\frac{19}{9\times9\times10\times10}\)
Chứng tỏ rằng:\(\frac{3}{1^2\times2^2}+\frac{5}{2^2\times3^2}+\frac{7}{3^2\times4^2}+...+\frac{39}{19^2\times20^2}\)< 1
\(\frac{3}{1^2x2^2}\)+\(\frac{5}{2^2x3^2}\)+...+\(\frac{39}{19^2x20^2}\)<1
=\(\frac{3}{1.4}\)+\(\frac{5}{4x9}\)+...+\(\frac{39}{361x400}\)<1
=1-\(\frac{1}{4}\)+\(\frac{1}{4}\)-...-\(\frac{1}{361}\)+\(\frac{1}{361}\)-\(\frac{1}{400}\)<1
vì 1-\(\frac{1}{400}\)<1 nên \(\frac{3}{1^2x2^2}\)+\(\frac{5}{2^2x3^2}\)+...+\(\frac{39}{39^2x40^2}\)<1
vậy..............................................
Tính nhanh : \(\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{9\times10}\)
\(\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{9\times10}\)
=\(2\times\frac{1}{1\times2}+2\times\frac{1}{2\times3}+2\times\frac{1}{3\times4}+...+2\times\frac{1}{9\times10}\)
=\(2\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\right)\)
=\(2\times\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
=\(2\times\left(\frac{1}{1}-\frac{1}{10}\right)=2\times\left(\frac{10}{10}-\frac{1}{10}\right)=2\times\frac{9}{10}\)
=\(\frac{9}{5}\)
=2-1+1-\(\frac{2}{3}\)+\(\frac{2}{3}\)-\(\frac{1}{2}\)+...+\(\frac{2}{9}\)-\(\frac{1}{5}\)
=2-\(\frac{1}{5}\)
=\(\frac{10}{5}\)-\(\frac{1}{5}\)
=\(\frac{9}{5}\).
**** mình nha mấy bạn.
Tìm x biết : \(\frac{2}{5}x\)+\(\frac{3}{10}=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\)
\(\frac{2}{5}x+\frac{3}{10}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.........+\frac{1}{9}-\frac{1}{10}\)
\(\frac{2}{5}x+\frac{3}{10}=1-\frac{1}{10}=\frac{9}{10}\)
\(\frac{2}{5}x=\frac{9}{10}-\frac{3}{10}=\frac{3}{5}\)
\(x=\frac{\frac{3}{5}}{\frac{2}{5}}=\frac{3}{2}\)
Ta có: \(\frac{1}{1x2}\)+ \(\frac{1}{2x3}\)+ \(\frac{1}{3x4}\)+ \(\frac{1}{4x5}\)+ .....+ \(\frac{1}{9x10}\)
= \(1-\left(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-.....-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
= 1 - \(\frac{1}{10}\)
= \(\frac{9}{10}\)
X = \(\frac{3}{2}\) là đúng đó bạn
tính tổng :\(B=\frac{5}{1\times2}+\frac{13}{2\times3}+\frac{25}{3\times4}+\frac{41}{4\times5}+...+\frac{181}{9\times10}\)
B=2+1/1.2+2+1/2.3+........+2+1/9.10
B=2.9+1/1.2+1/2.3+........+1/9.10
B=18+9/10
tính :\(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+\frac{1}{4\times5\times6}+\frac{1}{5\times6\times7}+\frac{1}{6\times7\times8}+\frac{1}{7\times8\times9}+\frac{1}{8\times9\times10}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)
\(\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{5\times6}\right)\times10-x=0\)
Ta thấy :
1/1x2 = 1/1 - 1/2
1/2x3 = 1/2 - 1/3
....
=>( 1/1x2 + 1/2x3 + 1/3x4 + 1/5x6 ) x 10 - x = ( 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 ) x 10 - x
= ( 1/1 - 1/6 ) x 10 - x =0
5/6 x 10 - x = 0
25/3 - x = 0
=> x = 25/3
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{5.6}\right).10-x=0\)
\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{30}\right).10-x=0\)
\(\left(1-\frac{1}{4}+\frac{1}{30}\right).10-x=0\)
\(\left(\frac{3}{4}+\frac{1}{30}\right).10-x=0\)
\(\frac{47}{60}.10-x=0\)
\(\frac{47}{6}-x=0\)
\(x=\frac{47}{6}-0\)
\(x=\frac{47}{6}\)
chứng minh
M=\(\dfrac{3}{1^2\times2^2}+\dfrac{5}{2^2\times3^2}+\dfrac{7}{3^2\times4^2}+.......+\dfrac{19}{9^2\times10^2}< 1\)
\(M=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(M=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}\)
\(M=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
\(M=1-\dfrac{1}{10^2}< 1\left(đpcm\right)\)