Question

cho ba số a,b,c thỏa mãn a+b+c =6 và \(\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}=\frac{3}{2}\).Tính giá trị của biểu thức \(P=\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c=a}\)

Answer 1

ban oi mk dat cau hoi nay cac ban giup mk vs

Answer 2

1/2x + 3/5 . ( x- 2 ) = 3

Answer 3

Ta có :

\(\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}=\frac{3}{2}\)

\(\Leftrightarrow\frac{c}{a+b}+1+\frac{b}{a+c}+1+\frac{a}{b+c}+1=\frac{3}{2}+3\)

\(\Leftrightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{a+c}+\frac{a+b+c}{b+c}=\frac{9}{2}\)

\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)=\frac{9}{2}\)

\(\Leftrightarrow6.\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)=\frac{9}{2}\)

\(\Rightarrow\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}=\frac{9}{2}:6=\frac{3}{4}\)

Vậy \(P=\frac{3}{4}\)