Question

Cho a > b > c biết : \(2\left(a^2+b^2\right)5ab\)

\(\left(a+b\right)^2=a^2+b^2+2ab\)

\(\left(a-b\right)^2=a^2+b^2-2ab\)

Tính \(\frac{3a-b}{2a+b}\)

Answer 1

Ta có:

\(2\left(a^2+b^2\right)=5ab\)

\(\Leftrightarrow2a^2-5ab+2b^2=0\)

\(\Leftrightarrow2a^2-4ab-ab+2b^2=0\)

\(\Leftrightarrow2a\left(a-2b\right)-b\left(a-2b\right)=0\)

\(\Leftrightarrow\left(2a-b\right)\left(a-2b\right)=0\)

\(\Leftrightarrow a=2b\) hay \(b=2a\)

Vì \(a>b>c\Leftrightarrow a=2b\)

\(\Leftrightarrow\frac{3a-b}{2a+b}=\frac{3.2b-b}{2.2b+b}=\frac{5b}{5b}=1\)

Vậy \(\frac{3a-b}{2a+b}=1\)