Question

Tìm x biết:

\(\dfrac{313-x}{101}+\dfrac{313-x}{103}+\dfrac{311-x}{105}+\dfrac{309-x}{107}=-4\)

Answer 1

Đề sai sửa lại và làm:

Ta có:

\(\dfrac{315-x}{101}+\dfrac{313-x}{103}+\dfrac{311-x}{105}+\dfrac{309-x}{107}=-4\)

\(\Leftrightarrow\left(\dfrac{315-x}{101}+1\right)+\left(\dfrac{313-x}{103}+1\right)+\left(\dfrac{311-x}{105}+1\right)+\left(\dfrac{309-x}{107}+1\right)=0\)

\(\Leftrightarrow\dfrac{416-x}{101}+\dfrac{416-x}{103}+\dfrac{416-x}{105}+\dfrac{416-x}{107}=0\)

\(\Leftrightarrow\left(416-x\right)\left(\dfrac{1}{101}+\dfrac{1}{103}+\dfrac{1}{105}+\dfrac{1}{107}\right)=0\)

\(\Leftrightarrow416-x=0\)

\(\Leftrightarrow x=416\)

VẬY....