Question

[Đỗ Anh Duy]

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Answer 1

[Nguyễn Ngọc Anh Minh]

1/

\(P^2=\frac{\left(a-b\right)^2}{\left(a+b\right)^2}=\frac{3\left(a^2+b^2-2ab\right)}{3\left(a^2+b^2+2ab\right)}=\frac{3a^2+3b^2-6ab}{3a^2+3b^2+6ab}=\frac{10ab-6ab}{10ab+6ab}=\frac{ab}{4}\)

Do \(a>b>0\Rightarrow P=\frac{a-b}{a+b}>0\)

\(\Rightarrow P=\frac{\sqrt{ab}}{2}\)

2/ Làm tương tự câu 1

Answer 2

[Nguyễn Ngọc Anh Minh]

Xin lỗi \(P^2=\frac{1}{4}\) Do \(a>b>0\Rightarrow P=\frac{a-b}{a+b}>0\)

\(\Rightarrow P=\frac{1}{2}\)