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ta có
(1/3+1/6+1/36) +(1/10+1/15+1/45)+(1/21+1/28)
=(\(\frac{12+6+1}{36}\)+\(\frac{9+6+2}{90}\)+\(\frac{4+3}{84}\)
19/36+17/90+1/12
=(19/36+1/12)+17/90
=7/12+17/90
=105/180+34/180
=139/180
1/3 +1/6+1/10+1/15+1/21+1/28+1/36+1/45
=1/1x3+1/3x2+1/2x5+1/3x5+1/3x7+1/7x4+1/4x9+1/9x5
=1/1-1/3+1/3-1/2....+1/9-1/5
=1/1
1/21 + 1/28 + 1/36 + ...+ 1/x(x+1)
=> 2/42 + 2/56 + 2/72 +....+ 2/x(x+1)
=> 2.(1/42 + 1/56 + 1/72 + ... + 1/x.(x+1))
=> 2 .(1/6.7 + 1/7.8 + 1/8.9 + ..+ 1/x.(x+1))
=> 2. ( 1/6 - 1/7 + 1/7-1/8 + ...+ 1/x - 1/x+1
=> 2 . (1/6 - 1/x+1)
=>1/3 - 2/x+1
\(\frac{1}{21}+\frac{1}{77}+\frac{1}{165}+...+\frac{1}{n^2+4n}=\frac{56}{673}\)
<=> \(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{n.\left(n+4\right)}=\frac{56}{673}\)
<=> \(4.\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{n.\left(n+4\right)}\right)=4.\frac{56}{673}\)
<=> \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{n\left(n+4\right)}=\frac{224}{673}\)
<=> \(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{n}-\frac{1}{n+4}=\frac{224}{673}\)
<=> \(\frac{1}{3}-\frac{1}{n+4}=\frac{224}{673}\)
<=> \(\frac{n+4-3}{3.\left(n+4\right)}=\frac{224}{673}\Leftrightarrow\frac{n}{3.\left(n+4\right)}=\frac{224}{673}\)
<=> 673n = 224.3(n+4)
<=> 673n = 224.3.n + 224.3.4
<=> 673n = 672n + 2688
<=> 673n - 672n = 2688
<=> n = 2688
* ĐK: \(x\ne0\)
Đề ra ...<=> \(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{9}\)
<=> \(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{2}{x\left(x+1\right)}=\frac{1}{9}\)
<=> \(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
<=>\(\frac{1}{6}-\frac{1}{x+1}+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
<=>\(\frac{1}{x+1}\left(1-\frac{1}{x}\right)=\frac{1}{6}-\frac{1}{9}\)
<=> \(\frac{x-1}{x\left(x+1\right)}=\frac{1}{36}\)
<=> \(\frac{x-1}{x\left(x-1\right)}=\frac{x-1}{36.\left(x-1\right)}\)
=> x(x-1) = 36. (x-1) => x =36
\(\frac{2}{2}.\left(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x+\left(x+1\right)}\right)=\frac{2}{9}\)
\(2\left(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2}{9}\)
\(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x.\left(x+1\right)}=\frac{1}{9}\)
\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{18}\)
x+1=18
x=18-1
x=17
\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-....+\frac{1}{256}-\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{512}\)
\(=\frac{255}{512}\)
Vậy \(A=\frac{255}{512}\)
=1/2-1/4+1/4-1/8+1/8-....+1/156-1/152
=1/2-1/152
=255/512
A=255/512
\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+....+\frac{1}{256}-\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{512}\)
\(=\frac{255}{512}\)
Vậy \(A=\frac{255}{512}\)
A=14 +18 +116 +132 +164 +1128 +1256 +1512
=12 −14 +14 −18 +....+1256 −1512
=12 −1512
=255512
Vậy A=255512
Phạm Long Khánh
Nhiều cách lắm,ví dụ nhé:
B = ( \(\frac{1}{4}+\frac{1}{5}+...+\frac{1}{11}\) ) + ( \(\frac{1}{12}+\frac{1}{13}+...+\frac{1}{19}\))
______________________ _________________________
B C
-Ta xét B ( vì bạn bảo chi tiết nên tôi làm như vậy còn ở bài thì không cần như vậy )
\(\frac{1}{4}>\frac{1}{12}\);...; \(\frac{1}{11}>\frac{1}{12}\)
-Xét C : \(\frac{1}{12}>\frac{1}{20};...;\frac{1}{19}>\frac{1}{20}\)
(=) B > \(\left(\frac{1}{12}+\frac{1}{12}+...+\frac{1}{12}\right)+\left(\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\right)\)
_________________ ___________________
8 số 8 số
(=) B > \(\frac{8}{12}+\frac{8}{20}\)= \(\frac{2}{3}+\frac{2}{5}\)= \(\frac{16}{15}\)> 1
(=) B > 1 (đpcm)
ngu thế. casio fx 570 mà lm
ta cần trình bày