*Tìm min:
\(P=\dfrac{a}{1-a}+\dfrac{b}{1-b}=\dfrac{1}{1-a}-1+\dfrac{1}{1-b}-1\)
\(\ge\dfrac{4}{\left(1-a\right)+\left(1-b\right)}-2\)
\(=\dfrac{4}{2-\dfrac{1}{2}}-2=\dfrac{2}{3}\)
Dấu "=" xảy ra khi \(a=b=\dfrac{1}{4}\). Do đó minP=2/3
*Tìm max: \(a,b\ge0\)
\(P=\dfrac{a}{1-a}+\dfrac{b}{1-b}=\dfrac{a-ab+b-ab}{\left(1-a\right)\left(1-b\right)}\)
\(=\dfrac{\dfrac{1}{2}-2ab}{1-\left(a+b\right)+ab}=\dfrac{\dfrac{1}{2}-2ab}{\dfrac{1}{2}+ab}=\dfrac{\dfrac{3}{2}-2\left(\dfrac{1}{2}+ab\right)}{\dfrac{1}{2}+ab}\)
\(=\dfrac{\dfrac{3}{2}}{\dfrac{1}{2}+ab}-2\le\dfrac{\dfrac{3}{2}}{\dfrac{1}{2}}-2=1\)
Dấu "=" xảy ra khi \(\left(a;b\right)=\left(0;\dfrac{1}{2}\right),\left(\dfrac{1}{2};0\right)\)
Vậy maxP=1