Question

[Ngô Thùy Linh]

Giải pt sau:

\(\dfrac{2032-x}{25}+\dfrac{2053-x}{23}+\dfrac{2070-x}{21}+\dfrac{2083-x}{19}=10\)

Answer 1

[Nguyễn Lê Phước Thịnh]

\(\Leftrightarrow\dfrac{2032-x}{25}-1+\dfrac{2053-x}{23}-2+\dfrac{2070-x}{21}-3+\dfrac{2083-x}{19}-4=0\)

=>2007-x=0

hay x=2007

Answer 2

[Nguyễn acc 2]

\(\dfrac{2032-x}{25}+\dfrac{2053-x}{23}+\dfrac{2070-x}{21}+\dfrac{2083-x}{19}=10\)

\(\Leftrightarrow\left(\dfrac{2032-x}{25}-1\right)+\left(\dfrac{2053-x}{23}-2\right)+\left(\dfrac{2070-x}{21}-3\right)+\left(\dfrac{2083-x}{19}-4\right)=0\)

\(\Leftrightarrow\dfrac{2032-x-25}{25}+\dfrac{2053-x-46}{23}+\dfrac{2070-x-63}{21}+\dfrac{2083-x-76}{19}=0\)

\(\Leftrightarrow\dfrac{2007-x}{25}+\dfrac{2007-x}{23}+\dfrac{2007-x}{21}+\dfrac{2007-x}{19}=0\)

\(\Leftrightarrow\left(2007-x\right)\left(\dfrac{1}{25}+\dfrac{1}{23}+\dfrac{1}{21}+\dfrac{1}{19}\right)=0\)

\(\Leftrightarrow2007-x=0\left(vì.\dfrac{1}{25}+\dfrac{1}{23}+\dfrac{1}{21}+\dfrac{1}{19}\ne0\right)\)

\(\Leftrightarrow x=2007\)