Question

tìm x

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

Answer 1

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

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

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

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

vì \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\ne0\)nên 416 - x = 0

\(\Rightarrow x=416\)

Answer 2

x=416