Question

a)Tim x    \(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+x=0\)

b)Tim a,b    (2a+5b+1)(2\(^a\)+a\(^2\)+a+b)=105    

Answer 1

\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+4=0\)

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

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

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

Do \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\ne0\)

=>416-x=0

<=>x=416