Cho \(S_1=1+\dfrac{1}{5}\)
\(S_2=1+\dfrac{1}{5}+\dfrac{1}{5^2}\)
................................
\(S_n=1+\dfrac{1}{5}+...+\dfrac{1}{5^n}\)
Chứng minh rằng: \(\dfrac{1}{5S_1^2}+\dfrac{1}{5^2S_2^2}+...+\dfrac{1}{5^nS_n^2}< \dfrac{1}{4}\)
cho số n nguyên dương và các tổng sau:
S\(_1\)=1+\(\dfrac{1}{5}\), S\(_2\)=1+\(\dfrac{1}{5}+\dfrac{1}{5^2}\), S\(_3\)=1+\(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}\), S\(_n\)=1+\(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+........+\dfrac{1}{5^n}\)
Chứng minh rằng: \(\dfrac{1}{5S_1^2}+\dfrac{1}{5^2S_2^2}+\dfrac{1}{5^3S^2_3}+.....+\dfrac{1}{5^nS^2_n}< \dfrac{35}{36}\)
chứng minh rằng :
a) \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\) b)\(\dfrac{1}{5^2}+\dfrac{1}{6^5}+...+\dfrac{1}{2013^2}+\dfrac{1}{2014}>\dfrac{1}{5}\)
Chứng minh rằng dãy số sau đây tăng và bị chặn trên :
\(x_1=\dfrac{1}{5+1};x_2=\dfrac{1}{5+1}+\dfrac{1}{5^2+1};x_3=\dfrac{1}{5+1}+\dfrac{1}{5^2+1}+\dfrac{1}{5^3+1},.....;x_n=\dfrac{1}{5+1}+\dfrac{1}{5^2+1}+.....+\dfrac{1}{5^n+1}\)
Chứng minh rằng :
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2013^2}+\dfrac{1}{2014^2}>\dfrac{1}{5}\)
Đặt \(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2014^2}\)
\(A>\dfrac{1}{5^2}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{2014.2015}\)
\(A>\dfrac{1}{5^2}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\)
\(A>\dfrac{1}{5^2}+\dfrac{1}{6}-\dfrac{1}{2015}\)
\(A>\dfrac{1}{5^2}+\dfrac{1}{6}-\dfrac{1}{150}=\dfrac{1}{5}\) (đpcm)
Chữ hơi xấu thông kẻm :>
Vội qá nên gạch xóa nhiều :>
1.tốc độ tb =\(\dfrac{s}{t}\)=\(\dfrac{s_1+s_2+...+s_n}{t_1+t_2+...+t_n}\)
2.vận tốc tb = độ dời/ khoảng tg=\(\dfrac{x_2-x_1}{t_2-t_1}\)
3.nửa quãng đg=\(\dfrac{2v_1v_2}{v_1+v_2}\)
4.nửa tg=\(\dfrac{v_1+v_2}{2}\)
5.1+2+3+...+n=\(\dfrac{n\left(n+1\right)}{2}\)
7.pt chuyển động x=x0+s=x0+vt
8.ptcđ của A: xa = xo + vat, B: xb = x'0 + vbt
Chứng minh rằng:
\(\dfrac{1}{3^2}+\dfrac{1}{5^2}+\dfrac{1}{7^2}+...+\dfrac{1}{99^2}< \dfrac{5}{18}\)
chứng minh rằng
A= \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 2\)
B=\(\dfrac{5}{11}+\dfrac{5}{12}+\dfrac{5}{13}+\dfrac{5}{14},1< B< 2\)
chứng minh :
a) \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4}\) b) \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2013^2}+\dfrac{1}{2014}>\dfrac{1}{5}\)
Cho A = \(\dfrac{\left(3\dfrac{2}{15}+\dfrac{1}{5}\right):2\dfrac{1}{2}}{\left(5\dfrac{3}{7}-2\dfrac{1}{4}\right):4\dfrac{43}{56}}\) ; B = \(\dfrac{1,2:\left(1\dfrac{1}{5}.1\dfrac{1}{4}\right)}{0,32+\dfrac{2}{25}}\)
Chứng minh rằng A= B
\(A=\dfrac{\left(3+\dfrac{2}{15}+\dfrac{1}{5}\right):\dfrac{5}{2}}{\left(5+\dfrac{3}{7}-2-\dfrac{1}{4}\right):\left(4+\dfrac{43}{56}\right)}\)
\(=\dfrac{\dfrac{10}{3}\cdot\dfrac{2}{5}}{\dfrac{89}{28}:\dfrac{267}{56}}=\dfrac{4}{3}:\dfrac{2}{3}=2\)
\(B=\dfrac{\dfrac{6}{5}:\left(\dfrac{6}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{8}{25}+\dfrac{2}{25}}=\dfrac{\dfrac{6}{5}:\dfrac{3}{2}}{\dfrac{2}{5}}=2\)
Do đó: A=B