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2/10+2/40+2/88+.....+2/340+2/460
=2/2.5+2/5.8+2/8.11+....+2/17.20+2/20.23
=2/3.(3/2.5+3/5.8+3/8.11+....+3/17.20+3/20.23)
=2/3.(1/2-1/5+1/5-1/8+1/8-1/11+.....+1/17-1/20+1/20-1/23)
=2/3.(1/2-1/23)=2/3.21/46=7/23
cho mình xin 1 ks
tách thành
=2/2x5+2/5x8+2/8x11+...+2/17x20+2/20x23
=2/3x(1+1/5-1/5+1/8-1/8+...+1/20-1/20+1/23)
=2/3x(1+1/23)
=2/3x24/23
=16/23
\(=\left(7-8\right)+\left(9-10\right)+...+\left(87-88\right)+89\)
\(=\left(-1\right)+\left(-1\right)+...+\left(-1\right)+89\)
Số các số hạng từ 7 đến 88 là: \(88-7+1=82\) (số hạng)
\(=\left(-1\right)\cdot\left(82\div2\right)+89\)
\(=\left(-1\right)\cdot41+89\)
\(=\left(-41\right)+89\)
\(=48\)
Sửa đề: B=11^87+1/11^88+1
\(11A=\dfrac{11^{90}+11}{11^{90}+1}=1+\dfrac{10}{11^{90}+1}\)
\(11B=\dfrac{11^{88}+11}{11^{88}+1}=1+\dfrac{10}{11^{88}+1}\)
mà 11^90>11^88
nên A<B
a) \(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
\(=1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}\)
\(=8-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(=8-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=8-\left(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+\frac{7-6}{6.7}+\frac{8-7}{7.8}+\frac{9-8}{8.9}+\frac{10-9}{9.10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)=7,6\)
b) Bạn làm tương tự.
\(=1+\frac{1}{10}+1+\frac{1}{40}+1+\frac{1}{88}+...+1+\frac{1}{460}\)
Nhưng mình chả hiểu nó viết theo qui luật gì ???