Question

a) Cho a + b +c = 2015 và \(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}=\frac{1}{2015}\)

Tính S = \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)

b) cho 2 số a,b thỏa mãn điều kiện a+b=1.Chứng minh a³ +b³ +ab lớn hơn hoặc bằng \(\frac{1}{2}\)

 

Answer 1

lớp 6 mà

Answer 2

lớp 9 đó

Answer 3

\(a)\) Ta có : 

\(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}=\frac{1}{2015}\)

\(\Leftrightarrow\)\(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)=\left(a+b+c\right).\frac{1}{2015}\)

\(\Leftrightarrow\)\(\frac{a+b+c}{a+b}+\frac{a+b+c}{a+c}+\frac{a+b+c}{b+c}=\frac{a+b+c}{2015}\)

\(\Leftrightarrow\)\(1+\frac{c}{a+b}+1+\frac{b}{a+c}+1+\frac{a}{b+c}=\frac{2015}{2015}\)

\(\Leftrightarrow\)\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=1-3\)

\(\Leftrightarrow\)\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=-2\)

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