cho c=5/5.8.11+5/8.11.14+5/302.305.308
c/m : c <1/48
Cho C= \(\dfrac{5}{5.8.11}+\dfrac{5}{8.11.14}+...+\dfrac{5}{302.305.308}\). Chứng minh C<\(\dfrac{1}{48}\)
\(C=\dfrac{5}{5\cdot8\cdot11}+\dfrac{5}{8\cdot11\cdot14}+...+\dfrac{5}{302\cdot305\cdot308}\\ =\dfrac{5}{6}\cdot\left(\dfrac{6}{5\cdot8\cdot11}+\dfrac{6}{8\cdot11\cdot14}+...+\dfrac{6}{302\cdot305\cdot308}\right)\\ =\dfrac{5}{6}\cdot\left(\dfrac{1}{5\cdot8}-\dfrac{1}{8\cdot11}+\dfrac{1}{8\cdot11}-\dfrac{1}{11\cdot14}+...+\dfrac{1}{302\cdot305}-\dfrac{1}{305\cdot308}\right)\\ =\dfrac{5}{6}\cdot\left(\dfrac{1}{40}-\dfrac{1}{305\cdot308}\right)\\ =\dfrac{5}{6}\cdot\dfrac{1}{40}-\dfrac{5}{6}\cdot\dfrac{1}{305\cdot308}\\ =\dfrac{1}{48}-\dfrac{5}{6\cdot305\cdot308}\\ \dfrac{5}{6\cdot305\cdot308}>0\Rightarrow\dfrac{1}{48}-\dfrac{5}{6\cdot305\cdot308}< \dfrac{1}{48}\)
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 B=5/5.8.11+5/8.11.14+...+5/302.305.308
Chứng minh B<1/48
Lời giải:
\(B=\frac{5}{5.8.11}+\frac{5}{8.11.14}+...+\frac{5}{302.205.308}\)
\(\Rightarrow \frac{6}{5}B=\frac{6}{5.8.11}+\frac{6}{8.11.14}+...+\frac{6}{302.305.308}\)
\(=\frac{11-5}{5.8.11}+\frac{14-8}{8.11.14}+...+\frac{308-302}{302.305.308}\)
\(=\frac{1}{5.8}-\frac{1}{8.11}+\frac{1}{8.11}-\frac{1}{11.14}+...+\frac{1}{302.305}-\frac{1}{305.308}\)
\(=\frac{1}{5.8}-\frac{1}{305.308}< \frac{1}{5.8}\)
\(\Rightarrow B< \frac{1}{40}.\frac{5}{6}\Leftrightarrow B< \frac{1}{48}\)
Cho C=\(\frac{5}{5.8.11}\)+ \(\frac{5}{8.11.14}\)+..............+\(\frac{5}{302.305.308}\)
Chứng minh C< \(\frac{1}{48}\)
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=\(\frac{5}{5.8.11}+\frac{5}{8.11.14}+...+\frac{5}{302.305.308}< \frac{1}{48}\)
Chứng minh rằng:
1)B=\(\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+...+\frac{3^{98}+1}{3^{98}}< 100\)
2)C=\(\frac{5}{5.8.11}+\frac{5}{8.11.14}+...+\frac{5}{302.305.308}\)<\(\frac{1}{48}\)
3)D=\(\frac{11}{9}+\frac{18}{16}+\frac{27}{25}+...+\frac{1766}{1764}\)
\(40\frac{20}{43}< D< 40\frac{20}{21}\)
Cau 1:1/1.6.11+1/6.11.16+1/11.16.21+...+1/96.101.106
Cau 2:5/3.7.11+5/7.11.15+5/11.15.19+...+5/91.95.99
Cau 3:1/1.2.3+1/2.3.4+1/3.4.5+...+1/98.99.100
Cau 4:5/2.5.8+5/5.8.11+5/8.11.14+...+5/95.98.101
Cau 5:2/1.5.9+2/5.9.13+2/9.13.17+2/13.17.21+2/17.21.25
Cau 6:4/2.6.10+4/6.10.14+4/10.14.18+...+4/94.98.102
Cau 7:3/5.10.15+3/10.15.20+3/15.20.25+...+3/90.95.100
Ban nao giai duoc het mk k cho
Minh dang can gap nhoa
C=5/58.11+5/8.11.14+3.6/4.5+...+5/302.305.308 CMR C<1/48
Cho D=5/5.8.11+5/11.14.17+...+5/302.305.308
CMR:D nho hon 1/48
Bạn chép đề sai rồi, mình sửa lại đề và làm luôn nhé :
Ta có :
\(D=\frac{5}{5.8.11}+\frac{5}{8.11.14}+...+\frac{5}{302.305.308}\)
\(\Rightarrow D=\frac{5}{6}.\left(\frac{6}{5.8.11}+\frac{6}{8.11.14}+...+\frac{6}{302.305.308}\right)\)
\(\Rightarrow D=\frac{5}{6}.\left(\frac{1}{5.8}-\frac{1}{8.11}+\frac{1}{8.11}\frac{1}{11.14}+...+\frac{1}{302.305}-\frac{1}{305.308}\right)\)
\(\Rightarrow D=\frac{5}{6}.\left(\frac{1}{5.8}-\frac{1}{305.308}\right)\)
\(\Rightarrow D=\frac{5}{6}.\frac{1}{40}-\frac{5}{6}.\frac{1}{305.308}\)
\(\Rightarrow D=\frac{1}{48}-\frac{5}{6.305.308}< \frac{1}{48}\) (đpcm )