\(A=\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+.+\dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}.\)So sánh A với 2
a) rút gọn: \(\dfrac{4^5x9^4-2x6^9}{2^{10}x3^8+6^8x20}\)
b) Cho A=\(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+\dfrac{4}{2^4}+\dfrac{5}{2^5}+...\dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}\).So sánh A với 2
a: \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
Cho S = \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+....+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) so sánh S và \(\dfrac{1}{5}\)
cho: A = \(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}\)so sánh A với 2
\(A=\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}\)
\(2A=1+1+\dfrac{3}{2^2}+...+\dfrac{99}{2^{98}}+\dfrac{100}{2^{99}}\)
\(2A-A=\left(1+1+\dfrac{3}{2^2}+...+\dfrac{99}{2^{98}}+\dfrac{100}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}\right)\)
\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)
Đặt:
\(B=1+\dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\)
\(2B=2+1+\dfrac{1}{2^2}+....+\dfrac{1}{2^{97}}+\dfrac{1}{2^{98}}\)
\(2B-B=\left(2+1+\dfrac{1}{2^2}+....+\dfrac{1}{2^{97}}+\dfrac{1}{2^{98}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\right)\)
\(B=2-\dfrac{1}{2^{99}}\)
Vậy \(A=2-\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}< 2\)
Bây giờ mình đang bận học bài 1 chút.Xíu nữa mình làm cho nhé
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 \(\dfrac{-1}{2}\)
2:
\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)
So sánh A với 1
\(A=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}\)
Ta có :
\(A=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+.....+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}\)
\(=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{3}}-\dfrac{1}{\sqrt{4}}+........+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}\) \(=1-\dfrac{1}{\sqrt{100}}< 1\)
Vậy \(A< 1\)
So sánh A và B :
\(A=1.3.5.7.....99\)
\(B=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}.....\dfrac{100}{2}\)
Lời giải:
\(A=1.3.5.7...99=\frac{1.2.3.4...99.100}{2.4.6.8.100}=\frac{1.2.3...99.100}{(1.2)(2.2)(3.2)...(50.2)}\)
\(=\frac{1.2.3...99.100}{(1.2.3...50).2^{50}}=\frac{51.52...100}{2^{50}}=\frac{51}{2}.\frac{52}{2}....\frac{100}{2}=B\)
Chứng minh rằng : A = \(\dfrac{1}{2}-\dfrac{2}{2^2}+\dfrac{3}{2^3}-\dfrac{4}{2^4}+....+\dfrac{99}{2^{99}}-\dfrac{100}{2^{100}}< \dfrac{2}{9}\)
Cho A = \(\dfrac{1}{2}x\dfrac{3}{4}x\dfrac{5}{6}x...x\dfrac{99}{100};B=\dfrac{1}{10}\) So sánh: A và B.
Tham khảo:
https://lazi.vn/edu/exercise/so-sanh-a-1-2-3-4-5-6-99-100-va-b-1-10
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<\(\dfrac{3}{16}\)
Lời giải:
\(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(3A=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(4A=A+3A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+....-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(12A=3-1+\frac{1}{3}-\frac{1}{3^2}+....-\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow 16A=12A+4A=3-\frac{101}{3^{99}}-\frac{100}{3^{100}}<3\)
\(\Rightarrow A< \frac{3}{16}\)