Lời giải:

\(B=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+....+\frac{2021}{4^{2021}}\)

\(4B=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2021}{4^{2020}}\)

\(4B-B=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2020}}-\frac{2021}{4^{2021}}\)

\(3B=1+\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{2020}}-\frac{2021}{4^{2021}}\)

\(12B=4+1+\frac{1}{4}+...+\frac{1}{4^{2019}}-\frac{2021}{4^{2020}}\)

\(9B=4-\frac{6067}{4^{2021}}<4\Rightarrow B< \frac{4}{9}< \frac{1}{2}\)

A=\(\dfrac{3}{1}\).(\(\dfrac{3}{2.5}\)+\(\dfrac{3}{5.8}\)+...+\(\dfrac{3}{98.101}\))

A=3.(\(\dfrac{1}{2}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{8}\)+...+\(\dfrac{1}{98}\)-\(\dfrac{1}{101}\))

A=3.(\(\dfrac{1}{2}\)-\(\dfrac{1}{101}\))

A=3.\(\dfrac{98}{202}\)

A=\(\dfrac{294}{202}\)=\(\dfrac{147}{101}\)

A = \(\dfrac{2022}{50^{10}}\) + \(\dfrac{2022}{50^8}\)

A = \(\dfrac{2022}{50^{10}}\) + \(\dfrac{2021}{50^8}\) +  \(\dfrac{1}{50^8}\)

B = \(\dfrac{2023}{50^{10}}\) + \(\dfrac{2021}{5^8}\) = \(\dfrac{2022}{50^{10}}\) + \(\dfrac{1}{50^{10}}\) + \(\dfrac{2021}{50^8}\)

Vì: \(\dfrac{1}{50^{10}}\) < \(\dfrac{1}{50^8}\) nên \(\dfrac{2022}{50^{10}}\) + \(\dfrac{2021}{50^8}\) + \(\dfrac{1}{50^{10}}\)  < \(\dfrac{2022}{50^{10}}\) + \(\dfrac{2021}{50^8}\) + \(\dfrac{1}{50^8}\)

Vậy A > B

ĐKXĐ: \(x>0\)

Áp dụng BĐT Cauchy cho 2 số dương:

\(\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\dfrac{1}{\sqrt{x}}}=2\)

Dấu "=" xảy ra \(\Leftrightarrow\left(\sqrt{x}\right)^2=1\Leftrightarrow x=1\left(tm\right)\)

mình tự bình loạn các bạn ạ

Ta đặt

  \(A=\dfrac{7}{1\times2}+\dfrac{7}{2\times3}+...+\dfrac{7}{99\times100}\)

\(\dfrac{1}{7}\times A=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+....+\dfrac{1}{99\times100}\)

\(\dfrac{1}{7}\times A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\dfrac{1}{7}\times A=1-\dfrac{1}{100}\)

\(\dfrac{1}{7}\times A=\dfrac{99}{100}\)

\(A=\dfrac{99}{100}\div\dfrac{1}{7}\)

\(A=\dfrac{693}{100}\)

= 7.(1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100)

= 7.(1 - 1/100)

= 7 . 99/100

= 693/100

\(A=7\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}=\)

\(=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{100-99}{99.100}=\)

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

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)

\(\Rightarrow A=7x\dfrac{99}{100}=6,93\)

GIUP VOI

HUHU

\(\Leftrightarrow x\cdot\dfrac{-1}{2}=-\dfrac{2}{5}\)

\(\Leftrightarrow x=\dfrac{2}{5}:\dfrac{1}{2}=\dfrac{4}{5}\)