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\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2+3^2}+...+\dfrac{19}{9^2-10^2}\)
\(=\) \(\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
\(=\) \(1-\dfrac{1}{10^2}< 1\) ( đpcm )
1:
a: =8/7-5/88=669/616
b: \(=1+\dfrac{2}{9}\cdot\dfrac{3}{7}-\dfrac{10}{7}=1+\dfrac{2}{21}-\dfrac{10}{7}\)
\(=\dfrac{21+2-30}{21}=\dfrac{-7}{21}=\dfrac{-1}{3}\)
c: \(=\dfrac{11}{3}-\dfrac{6}{7}+4=\dfrac{77-18+84}{21}=\dfrac{143}{21}\)
Bài 2:
a: =>9/4-x=5/11*2=10/11
=>x=9/4-10/11=59/44
b: =>2/9:x=19/21
=>x=2/9:19/21=14/57
b)=3^1+(3^2+3^3+3^4)+(3^5+3^6+3^7)+....+(3^58+3^59+3^60)
=3^1+(3^2.1+3^2.3+3^2.9)+(3^5.1+3^5.3+3^5.9)+......+(3^58.1+3^58.3+3^58.9)
=3^1+3^2.(1+3+9)+3^5.(1+3+9)+.....+3^58.(1+3+9)
=3+3^2.13+3^5.13+.........+3^58.13
=3.13.(3^2+3^5+....+3^58)
vi tich tren co thua so 13 nen tich do chia het cho 13
=
bai1
a) A=(31+32)+(33+34)+...+(359+360)
=(3^1.1+3^1.3)+...+(3^59.1+3^59.2)
=3^1.(1+3)+...+3^59.(1+3)
=3^1.4+....+3^59.4
=4.(3^1+...+3^59)
vi tich tren co thua so 4 nen tich do chia het cho 4
a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)
b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)
\(\frac{3}{1^2\cdot2^2}+\frac{5}{2^2\cdot3^2}+...+\frac{19}{9^2\cdot10^2}\)\(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{9^2}-\frac{1}{10^2}=1-\frac{1}{10^2}=\frac{99}{100}\)<1
ai giúp mình với rồi mình tink cho nha cảm ơn các bạn nhiều