Tính tổng S và so sánh với 0,1 biết : S= 1/10+1/40+1/88+1/154+1/238+1/340
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S = \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)
S = \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)
S = \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{17}-\frac{1}{20}\right)\)
S = \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{1}{3}.\frac{9}{20}\)
S = \(\frac{3}{20}\)
S = 0,15 > 0,1
a) A = 1/2.5 + 1/5.8 + 1/8.11 + 1/11.14 + 1/14.17 + 1/17.20
=> 3A = 1/2 - 1/5 + 1/5 - .... + 1/14 - 1/17 + 1/17 - 1/20
=> 3A = 1/2 - 1/20 = 9/20
=> A = 3/20
b) 200410 + 20049 = 20049(1+2004) = 20049 . 2005
200510 = 20059 . 2005
Do 20059 > 20049 nên 200410 + 20049 < 200510
\(A=\frac{3}{3}.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\right)\)
\(A=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{3}{20}\)
S=1/2.5 + 1/5.8 + 1/8.11 + 1/11.14 + 1/14.17 + 11/17.20
S=1/3.(1/2-1/5+1/5-1/8+.......+1/17 - 1/20)
S=1/3.(1/2-1/20)
S=1/3.9/20
S=3/20
Ai tích mk mk sẽ tích lại
Chú ý tích gấp ddooi khi tích đủ 3 cái
Ta có: S = 1/10 + 1/40 + 1/88 + 1/154 + 1/238 + 1/340 + 1/460 + 1/598 + 1/754 + 1/928
=> S = 1/2.5 + 1/5.8 + 1/8.11 + ... + 1/26.29 + 1/29.32
Nhân 2 vế với 3 và áp dụng công thức tách 1 phân số thành hiệu 2 phân số: x/n.(n + x) = 1/n - 1/(n + x)
=> 3.S = 3.(1/2.5 + 1/5.8 + 1/8.11 + ... + 1/26.29 +1/29.32)
=> 3.S = 3/2.5 + 3/5.8 + 3/8.11 + ... + 3/26.29 + 3/29.32
=> 3.S = 1/2 - 1/ 5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/26 - 1/29 + 1/29 - 1/32
=> 3.S = 1/2 - 1/32
=> 3.S = 15/32
=> S = 5/32
@Cre: G+
S=1/2.5 + 1/5.8 + 1/8.11 + 1/11.14 + 1/14.17 + 11/17.20
S=1/3.(1/2-1/5+1/5-1/8+.......+1/17 - 1/20)
S=1/3.(1/2-1/20)
S=1/3.9/20
S=3/20
\(A=\frac{9}{10}+\frac{39}{40}+\frac{87}{88}+\frac{153}{154}+\frac{237}{238}+\frac{339}{340}=\frac{117}{20}\)
\(suyra:A<1\)
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)
\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{20}\right)\)
\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{3}{20}\)
\(S=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{184}+\frac{1}{238}+\frac{1}{340}=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{1}{3}.\frac{9}{20}=\frac{3}{20}>\frac{2}{20}=\frac{1}{10}=0,1\)
vậy S>0,1