các bạn giúp mình với ạ:3
\(P=\dfrac{\left(\dfrac{a}{b}+\dfrac{b}{a}+1\right)\left(\dfrac{1}{a}-\dfrac{1}{b}\right)^2}{\left(\dfrac{a}{b}+\dfrac{b}{a}\right)^2-\left(\dfrac{a}{b}+\dfrac{b}{a}\right)-2}=\dfrac{\left(\dfrac{a}{b}+\dfrac{b}{a}+1\right)\left(\dfrac{1}{a}-\dfrac{1}{b}\right)^2}{\left(\dfrac{a}{b}+\dfrac{b}{a}+1\right)\left(\dfrac{a}{b}+\dfrac{b}{a}-2\right)}\)
\(=\dfrac{\left(\dfrac{1}{a}-\dfrac{1}{b}\right)^2}{\dfrac{a}{b}+\dfrac{b}{a}-2}=\dfrac{\left(\dfrac{a-b}{ab}\right)^2}{\dfrac{a^2+b^2-2ab}{ab}}=\dfrac{\dfrac{\left(a-b\right)^2}{a^2b^2}}{\dfrac{\left(a-b\right)^2}{ab}}=\dfrac{ab}{a^2b^2}=\dfrac{1}{ab}\)
b.
\(4a+b+\sqrt{ab}=1\Leftrightarrow4a-4\sqrt{ab}+b+5\sqrt{ab}=1\)
\(\Leftrightarrow\left(2\sqrt{a}-\sqrt{b}\right)^2+5\sqrt{ab}=1\)
\(\Leftrightarrow5\sqrt{ab}\le1\)
\(\Leftrightarrow\dfrac{1}{\sqrt{ab}}\ge5\)
\(\Leftrightarrow\dfrac{1}{ab}\ge25\)
\(\Rightarrow P_{min}=25\) khi \(\left(a;b\right)=\left(\dfrac{1}{10};\dfrac{2}{5}\right)\)
