Câu hỏi của Nguyễn Thị Ngọc Ly

Question

So sánh A=2021/2022 và B=1/1.2+1/2.3+1/3.4+2...+1/97.98+1/98.99

Đoàn Đức Hà · Accepted Answer

$B=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}$

$=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{99-98}{98.99}$

$=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}$

$=1-\dfrac{1}{99}$

$A=\dfrac{2021}{2022}=\dfrac{2022-1}{2022}=1-\dfrac{1}{2022}$

Có $2022>99>0\Leftrightarrow\dfrac{1}{99}>\dfrac{1}{2022}$

Suy ra $A>B$.Câu trả lời: $B=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}$

$=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{99-98}{98.99}$

$=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}$

$=1-\dfrac{1}{99}$

$A=\dfrac{2021}{2022}=\dfrac{2022-1}{2022}=1-\dfrac{1}{2022}$

Có $2022>99>0\Leftrightarrow\dfrac{1}{99}>\dfrac{1}{2022}$

Suy ra $A>B$.