Question

1)   Cho 3 số a,b,c,t/m: abc=1

Tính \(S=\frac{1}{1+a+ab}+\frac{1}{1+b+bc}+\frac{1}{1+c+ac}\)

2)  Cho a>b, so sánh \(\frac{a+c}{b+c}\)và \(\frac{a}{b}\)

 

Answer 1

\(S=\frac{1}{1+a+ab}+\frac{1}{1+b+bc}+\frac{1}{1+c+ca}\)

\(\Rightarrow S=\frac{abc}{abc+a+ab}+\frac{1}{1+b+bc}+\frac{abc}{abc+c.abc+ca}\)

\(S=\frac{abc}{a.\left(bc+b+1\right)}+\frac{1}{1+b+bc}+\frac{abc}{ac.\left(bc+b+1\right)}\)

\(S=\frac{bc}{bc+b+1}+\frac{1}{1+b+bc}+\frac{b}{bc+b+1}\)

\(S=\frac{bc+b+1}{bc+b+1}\)

\(S=1\)

Điều kiện \(c\ge0\);\(a;b>0\)

Ta có: \(a>b\)

\(\Rightarrow ac\ge bc\)

\(\Rightarrow ac+ab\ge bc+ab\)

\(a.\left(b+c\right)\ge b.\left(c+a\right)\)

\(\Rightarrow\frac{a+c}{b+c}\ge\frac{a}{b}\)

Tham khảo nhé~