Question

tìm x,biết:  \(\frac{45-x}{1963}+\frac{40-x}{1968}+\frac{35-x}{1973}+\frac{30-x}{1978}+4=0\)

Answer 1

Ta có : \(\frac{45-x}{1963}+\frac{40-x}{1968}+\frac{35-x}{1973}+\frac{30-x}{1978}+4=0\)

\(\Leftrightarrow\frac{45-x}{1963}+1+\frac{40-x}{1968}+1+\frac{35-x}{1973}+1+\frac{30-x}{1978}=0\)

\(\Leftrightarrow\frac{2008-x}{1963}+\frac{2008-x}{1968}+\frac{2008-x}{1973}+\frac{2008-x}{1978}=0\)

\(\Leftrightarrow\left(2008-x\right)\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)=0\)

Vì \(\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)\ne0\)

Nên : 2008 - x = 0 

<=> x = 2008

Vậy x = 2008