Đáp án: $x=20$
Giải thích các bước giải:
Ta có:
$A=\dfrac{1}{1.201}+\dfrac{1}{2.202}+...+\dfrac{1}{10.210}$
$\to 200A=\dfrac{200}{1.201}+\dfrac{200}{2.202}+...+\dfrac{200}{10.210}$
$\to 200A=\dfrac{201-1}{1.201}+\dfrac{202-2}{2.202}+...+\dfrac{210-10}{10.210}$
$\to 200A=\dfrac11-\dfrac1{201}+\dfrac12-\dfrac1{202}+...+\dfrac1{10}-\dfrac{1}{210}$
$\to 200A=\left(1+\dfrac12+..+\dfrac1{10}\right)-\left(\dfrac{1}{201}+\dfrac{1}{202}+...+\dfrac{1}{210}\right)$
$\to A=\dfrac{1}{200}\left(\left(1+\dfrac12+..+\dfrac1{10}\right)-\left(\dfrac{1}{201}+\dfrac{1}{202}+...+\dfrac{1}{210}\right)\right)$
Lại có:
$B=\dfrac{1}{1.11}+\dfrac{1}{2.12}+...+\dfrac{1}{200.210}$
$\to 10B=\dfrac{10}{1.11}+\dfrac{10}{2.12}+...+\dfrac{10}{200.210}$
$\to 10B=\dfrac{11-1}{1.11}+\dfrac{12-2}{2.12}+...+\dfrac{210-200}{200.210}$
$\to 10B=\dfrac11-\dfrac1{11}+\dfrac12-\dfrac1{12}+...+\dfrac1{200}-\dfrac1{210}$
$\to 10B=\left(1+\dfrac12+...+\dfrac1{200}\right)-\left(\dfrac1{11}+\dfrac{12}+...+\dfrac1{210}\right)$
$\to 10B=\left(1+\dfrac12+..+\dfrac1{10}\right)+\left(\dfrac1{11}+\dfrac1{12}+...+\dfrac1{200}\right)-\left(\dfrac1{11}+\dfrac1{12}+...+\dfrac1{200}\right)-\left(\dfrac1{201}+\dfrac1{202}+...+\dfrac{1}{210}\right)$
$\to 10B=\left(1+\dfrac12+..+\dfrac1{10}\right)-\left(\dfrac1{201}+\dfrac1{202}+...+\dfrac{1}{210}\right)$
$\to B=\dfrac1{10}\left(\left(1+\dfrac12+..+\dfrac1{10}\right)-\left(\dfrac1{201}+\dfrac1{202}+...+\dfrac{1}{210}\right)\right)$
Khi đó phương trình trở thành:
$\dfrac{1}{200}\left(\left(1+\dfrac12+..+\dfrac1{10}\right)-\left(\dfrac{1}{201}+\dfrac{1}{202}+...+\dfrac{1}{210}\right)\right)\cdot x=\dfrac1{10}\left(\left(1+\dfrac12+..+\dfrac1{10}\right)-\left(\dfrac1{201}+\dfrac1{202}+...+\dfrac{1}{210}\right)\right)$
$\to x=20$
