a: \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)

\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\)

\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)

\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)

\(A=\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}\)

\(2A=1+1+\dfrac{3}{2^2}+...+\dfrac{99}{2^{98}}+\dfrac{100}{2^{99}}\)

\(2A-A=\left(1+1+\dfrac{3}{2^2}+...+\dfrac{99}{2^{98}}+\dfrac{100}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}\right)\)

\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)

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

2:

\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)

\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)

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 \(A< 1\)

Lời giải:

\(A=1.3.5.7...99=\frac{1.2.3.4...99.100}{2.4.6.8.100}=\frac{1.2.3...99.100}{(1.2)(2.2)(3.2)...(50.2)}\)

\(=\frac{1.2.3...99.100}{(1.2.3...50).2^{50}}=\frac{51.52...100}{2^{50}}=\frac{51}{2}.\frac{52}{2}....\frac{100}{2}=B\)

B>A?

Tham khảo:

https://lazi.vn/edu/exercise/so-sanh-a-1-2-3-4-5-6-99-100-va-b-1-10

\(B>A\)

Lời giải:

\(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

\(3A=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)

\(4A=A+3A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+....-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(12A=3-1+\frac{1}{3}-\frac{1}{3^2}+....-\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

\(\Rightarrow 16A=12A+4A=3-\frac{101}{3^{99}}-\frac{100}{3^{100}}<3\)

\(\Rightarrow A< \frac{3}{16}\)