\(a,x=81\Leftrightarrow\sqrt{x}=9\Leftrightarrow A=\dfrac{9-5}{9}=\dfrac{4}{9}\\ b,P=A\cdot B=\dfrac{\sqrt{x}-5}{\sqrt{x}}\cdot\dfrac{x+5\sqrt{x}-3\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ P=\dfrac{\sqrt{x}-5}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}+5}\)
\(c,P^2-P=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+5}\right)^2-\dfrac{\sqrt{x}+2}{\sqrt{x}+5}\\ =\dfrac{\sqrt{x}+2}{\sqrt{x}+5}\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+5}-1\right)=\dfrac{\sqrt{x}+2}{\sqrt{x}+5}\cdot\dfrac{-3}{\sqrt{x}+5}=\dfrac{-3\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+5\right)^2}\)
Ta có \(\left(\sqrt{x}+5\right)^2>0;-3\left(\sqrt{x}+2\right)< 0\left(-3< 0;\sqrt{x}+2>0\right)\)
Do đó \(P^2-P< 0\Leftrightarrow P^2< P\)
Em cần giúp câu c và d ạ, mn giúp em với em đang cần gấp
