Xét \(3a^2+3b^2=10ab\Rightarrow a^2+b^2=\frac{10ab}{3}\)
hay: \(a^2+b^2=\frac{10}{3}ab\Rightarrow a^2+b^2+2ab=\frac{10}{3}ab+2ab\Rightarrow\left(a+b\right)^2=\frac{16}{3}ab\) (1)
\(a^2+b^2=\frac{10}{3}ab\Rightarrow a^2+b^2-2ab=\frac{10}{3}ab-2ab\Rightarrow\left(a-b\right)^2=\frac{4}{3}ab\) (2)
Ta có \(p=\frac{a+b}{a-b}\Rightarrow p^2=\frac{\left(a+b\right)^2}{\left(a-b\right)^2}=\frac{\frac{16}{3}ab}{\frac{4}{3}ab}=4\) Vậy \(p=2\) hoặc \(p=-2\)
ta có 3a^2 +3b^2=10ab
<=> 3a(a-3b) - b(a-3b)=0
<=> (3a-b)(a-3b)=0
=> a=3b ; 3a=b (loại vì a>b>0)
thay a=3b
ta có P=3b-b/3a+b
= 2b/4b
=1/2
