a) b=1/3+1/15+1/35+...+1/97.99
b) c=2/1.2.3+2/2.3.4+2/3.4.5+...+2/98.99.100
c) d=5/2.3.4+5/3.4.5+...+5/98.99.100+5/99.100.101
GIẢI GIÚP MÌNH THEO CÁCH HỌC CỦA LỚP 6 VỚI Ạ. CẢM ƠN MỌI NGƯỜI NHIỀU!
a/
\(b=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)
\(2b=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{99-97}{97.99}=\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}=\)
\(=1-\dfrac{1}{99}=\dfrac{98}{99}\Rightarrow b=\dfrac{98}{2.99}=\dfrac{49}{99}\)
b/
\(c=\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}=\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+\dfrac{1}{98.99}-\dfrac{1}{99.100}=\)
\(=\dfrac{1}{2}-\dfrac{1}{99.100}\)
c/
\(\dfrac{2}{5}.d=\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}+\dfrac{101-99}{99.100.101}=\)
\(=\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}+\dfrac{1}{99.100}-\dfrac{1}{100.101}=\)
\(=\dfrac{1}{2.3}-\dfrac{1}{100.101}\Rightarrow d=\left(\dfrac{1}{2.3}-\dfrac{1}{100.101}\right):\dfrac{2}{5}\)
Cho B=5/5.8.11+5/8.11.14+...+5/302.305.308
Chứng minh B<1/48
\(B=\frac{5}{5.8.11}+\frac{5}{8.11.14}+...+\frac{5}{302.305.308}\)
\(\Rightarrow\frac{6}{5}B=\frac{6}{5.8.11}+\frac{6}{8.11.14}+...+\frac{6}{302.305.308}\)
\(=\frac{11-5}{5.8.11}+\frac{14-8}{8.11.14}+...+\frac{308-302}{302.305.308}\)
\(=\frac{1}{5.8}-\frac{1}{8.11}+\frac{1}{8.11}-\frac{1}{8.11}+...+\frac{1}{302.305}-\frac{1}{305.308}\)
\(=\frac{1}{5.8}-\frac{1}{305.308}< \frac{1}{5.8}\)
cho c=5/5.8.11+5/8.11.14+5/302.305.308
c/m : c <1/48
Cho B=5/5.8.11+5/8.11.14+...+5/302.305.308
Chứng minh B<1/48
Lời giải:
\(B=\frac{5}{5.8.11}+\frac{5}{8.11.14}+...+\frac{5}{302.205.308}\)
\(\Rightarrow \frac{6}{5}B=\frac{6}{5.8.11}+\frac{6}{8.11.14}+...+\frac{6}{302.305.308}\)
\(=\frac{11-5}{5.8.11}+\frac{14-8}{8.11.14}+...+\frac{308-302}{302.305.308}\)
\(=\frac{1}{5.8}-\frac{1}{8.11}+\frac{1}{8.11}-\frac{1}{11.14}+...+\frac{1}{302.305}-\frac{1}{305.308}\)
\(=\frac{1}{5.8}-\frac{1}{305.308}< \frac{1}{5.8}\)
\(\Rightarrow B< \frac{1}{40}.\frac{5}{6}\Leftrightarrow B< \frac{1}{48}\)
Chứng minh :
1,C=\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{70}.C< \frac{3}{4}\)
2,D=\(\frac{1}{5^2}+\frac{1}{9^2}+...+\frac{1}{409^2}< \frac{1}{12}\)
3,E=\(\frac{5}{5.8.11}+\frac{5}{8.11.14}+...+\frac{5}{302.305.308}< \frac{1}{48}\)
1) Tìm x biết:
\(\left(1-\dfrac{3}{10}-x\right):\left(\dfrac{19}{10}-1-\dfrac{2}{5}\right)+\dfrac{4}{5}=1\)
2) Tính nhanh
a)\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{10.11.12}\)
b)\(\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^2}{3.4}.\dfrac{4^2}{4.5}\)
câu b bài 2:
\(\dfrac{1^2}{1\cdot2}\cdot\dfrac{2^2}{2\cdot3}\cdot\dfrac{3^2}{3\cdot4}\cdot\dfrac{4^2}{4\cdot5}\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\)
\(=\dfrac{1}{5}\)
câu a bài 2:
\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{10\cdot11\cdot12}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}-...-\dfrac{1}{12}\)
\(=1-\dfrac{1}{12}=\dfrac{11}{12}\)
Cau 1: Tim n biet : 5/8 + 5/24 + 5/48 + 5/80 + ... + 5/2n + 2 . 2n + 4 = 189/112
Cau 2 : Cho A = 1 + 1/1.2 + 1/1.2.3 + ... + 1/1.2.3...2014. So sanh A voi 2
Cau 3 : Tim n biet : 5/3 + 5/15 + 5/35 + 5/63 +...+ 5/2n + 1 . 2n + 3 = 172/69
bài 2 : s = 4 . 5 + 5.6 + 6.7+...+ 100.101
bài 3 : s= 1.2.3 + 2.3.4+ 3.4.5+...+ 98 . 99 .100
bài 4 : tính tổng sau: 1/5 + 1/25 + ... + 1/5 mu 100
Bài 1:
$A=1.2+2.3+3.4+...+201.202$
$3A=1.2.3+2.3(4-1)+3.4(5-2)+....+201.202(203-200)$
$=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+201.202.203-200.201.202$
$=(1.2.3+2.3.4+3.4.5+...+201.202.203)-(1.2.3+2.3.4+....+200.201.202)$
$=201.202.203$
$\Rightarrow A=\frac{201.202.203}{3}=2747402$
Bài 2:
$S=4.5+5.6+6.7+....+100.101$
$3S=4.5(6-3)+5.6.(7-4)+6.7.(8-5)+....+100.101(102-99)$
$=4.5.6-3.4.5+5.6.7-4.5.6+6.7.8-5.6.7+....+100.101.102-99.100.101$
$=(4.5.6+5.6.7+6.7.8+...+100.101.102)-(3.4.5+4.5.6+5.6.7+...+99.100.101)$
$=100.101.102-3.4.5$
$\Rightarrow S=\frac{100.101.102-3.4.5}{3}=343380$
Bài 3:
$S=1.2.3+2.3.4+3.4.5+...+98.99.100$
$4S=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+...+98.99.100(101-97)$
$=(1.2.3.4+2.3.4.5+3.4.5.6+...+98.99.100.101)-(0.1.2.3+1.2.3.4+2.3.4.5+...+97.98.99.100)$
$=98.99.100.101$
$\Rightarrow S=\frac{98.99.100.101}{4}$
C= 1/3.7.11+1/7.11.15+1/11.15.19+1/15.19.23+1/19.23.27+1/23.27.31
8C=8/3.7.11+8/7.11.15+8/11.15.19+8/15.19.23+8/19.23.27+8/23.27.31
8C=1/3.7-1/7.11+1/7.11-1/11.15+1/11.15-1/15.19+1/15.19-1/19.23+1/19.23+-1/23.27-1/27.31
8C=1/3.7-1/27.31
8C=1/21-1/837
8C=272/5859
C=34/5859
