Tìm `x` `(x-20) · (1/2 + 1/3 + 1/4 + ... + 1/200)/(1/199 + 2/198 + ... + 199/1) = 1/2000` Nhanh nha mng !!

Đáp án: $x=\dfrac{201}{10}$

Giải thích các bước giải:

$\left( x-20 \right)\cdot \dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{200}}{\dfrac{1}{199}+\dfrac{2}{198}+...+\dfrac{199}{1}}=\dfrac{1}{2000}$

Xét mẫu: $\dfrac{1}{199}+\dfrac{2}{198}+...+\dfrac{199}{1}$

$=1+\left( \dfrac{1}{199}+1 \right)+\left( \dfrac{2}{198}+1 \right)+...+\left( \dfrac{198}{2}+1 \right)$

$=\dfrac{200}{200}+\dfrac{200}{199}+\dfrac{200}{198}+...+\dfrac{200}{2}$

$=200\left( \dfrac{1}{200}+\dfrac{1}{199}+\dfrac{1}{198}+...+\dfrac{1}{2} \right)$

Vậy: $\left( x-20 \right)\cdot \dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{200}}{\dfrac{1}{199}+\dfrac{2}{198}+...+\dfrac{199}{1}}=\dfrac{1}{2000}$

$\to \left( x-20 \right)\cdot \dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{200}}{200\left( \dfrac{1}{200}+\dfrac{1}{199}+\dfrac{1}{198}+...+\dfrac{1}{2} \right)}=\dfrac{1}{2000}$

$\to \left( x-20 \right)\cdot \dfrac{1}{200}=\dfrac{1}{2000}$

$\to x-20=\dfrac{1}{2000}:\dfrac{1}{200}$

$\to x-20=\dfrac{1}{10}$

$\to x=\dfrac{1}{10}+20$

$\to x=\dfrac{201}{10}$