Cho a+b+c=20170 và 1/(a+b) + 1/(a+c) + 1/(b+c)=1/10 Tính S= a/(b+c) + b/(a+c) + c/(a+b) 05/09/2021 Bởi Aubrey Cho a+b+c=20170 và 1/(a+b) + 1/(a+c) + 1/(b+c)=1/10 Tính S= a/(b+c) + b/(a+c) + c/(a+b)
Có a+b+c = 20170 => a= 20170- (b+c) b= 20170 – (a+c) c= 20170 – (a+b) S = $\frac{a}{b+c}$ + $\frac{b}{a+c}$ +$\frac{c}{a+b}$ = $\frac{20170- (b+c)}{b+c}$ + $\frac{ 20170 – (a+c)}{a+c}$+$\frac{20170 – (a+b)}{a+b}$ = $\frac{20170}{b+c}$- 1+ $\frac{20170}{a+c}$- 1+ $\frac{20170}{a+b}$- 1 = 20170( $\frac{1}{b+c}$ + $\frac{1}{a+c}$ +$\frac{1}{a+b}$)-3 = 20170. $\frac{1}{10}$- 3 = 2017 -3= 2014 Bình luận
Có a+b+c = 20170
=> a= 20170- (b+c)
b= 20170 – (a+c)
c= 20170 – (a+b)
S = $\frac{a}{b+c}$ + $\frac{b}{a+c}$ +$\frac{c}{a+b}$
= $\frac{20170- (b+c)}{b+c}$ + $\frac{ 20170 – (a+c)}{a+c}$+$\frac{20170 – (a+b)}{a+b}$
= $\frac{20170}{b+c}$- 1+ $\frac{20170}{a+c}$- 1+ $\frac{20170}{a+b}$- 1
= 20170( $\frac{1}{b+c}$ + $\frac{1}{a+c}$ +$\frac{1}{a+b}$)-3
= 20170. $\frac{1}{10}$- 3
= 2017 -3= 2014