Các câu hỏi tương tự
1/ Cho A dfrac{1}{3}-dfrac{2}{3^2}+dfrac{3}{3^3}-dfrac{4}{3^4}+.....+dfrac{99}{3^{99}}-dfrac{100}{3^{100}} Chứng minh A dfrac{3}{16}2/ Cho B(dfrac{1}{2^2}-1)(dfrac{1}{3^2}-1)....(dfrac{1}{100^2}-1) So sánh B và dfrac{-1}{2}
Đọc tiếp
1/ Cho A= \(\dfrac{1}{3}\)-\(\dfrac{2}{3^2}\)+\(\dfrac{3}{3^3}\)-\(\dfrac{4}{3^4}\)+.....+\(\dfrac{99}{3^{99}}\)-\(\dfrac{100}{3^{100}}\) Chứng minh A < \(\dfrac{3}{16}\)
2/ Cho B=(\(\dfrac{1}{2^2}\)-1)(\(\dfrac{1}{3^2}\)-1)....(\(\dfrac{1}{100^2}\)-1) So sánh B và \(\dfrac{-1}{2}\)
Chứng minh: M= \(\dfrac{1}{3}\)+ \(\dfrac{2}{3^2}\)+ \(\dfrac{3}{3^3}\) + .....+ \(\dfrac{100}{3^{100}}\) <\(\dfrac{3}{4}\)
Chứng minh rằng : \(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>10\)
1/ Cho A= \(\dfrac{1}{3}\)-\(\dfrac{2}{3^2}\)+\(\dfrac{3}{3^3}\)-\(\dfrac{4}{3^4}\)+...+\(\dfrac{99}{3^{99}}\)-\(\dfrac{100}{3^{100}}\)
c/m A<\(\dfrac{3}{16}\)
20 tháng 9 2023 lúc 15:50
Cho biểu thức : \(C=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}+\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) CMR: \(C< \dfrac{3}{16}\)
CMR: \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}< \dfrac{3}{16}\)
9 tháng 1 lúc 20:34
\(\dfrac{3}{4}+\dfrac{1}{4}:x=-2\dfrac{1}{2}\)
\(\left(\dfrac{2}{3}\right)^{100}:x=\left(\dfrac{-2}{3}\right)^{98}\)
\(\dfrac{3}{2}:\left|4x-\dfrac{1}{5}\right|=\dfrac{3}{4}\)
a, \(P=\dfrac{1+2}{1^2\cdot2^2}+\dfrac{2+3}{2^2\cdot3^2}+...+\dfrac{9+10}{9^2\cdot10^2}\)
Chứng minh rằng: \(P< 1\)
b, \(Q=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)
Chứng minh rằng: \(Q< \dfrac{1}{2}\)
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Mình cần gấp lắm. 13/08/2017 là mình cần rồi.
B=(\(\dfrac{3}{2}-\dfrac{2}{2^2}\))\(\times\)(\(\dfrac{4}{3}-\dfrac{2}{3^2}\))\(\times\)...\(\times\)(\(\dfrac{101}{100}-\dfrac{2}{100^2}\))