\(P=\frac{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}-2\right)}{\sqrt{x}+1}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\sqrt{x}+1}\)
\(P=\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=x-3\sqrt{x}+2\)
\(P=\left(\sqrt{x}-\frac{3}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
\(P_{Min}=-\frac{1}{4}\) khi \(\sqrt{x}=\frac{3}{2}\Leftrightarrow x=\frac{9}{4}\)
b/ \(Q=\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(-x+3\sqrt{x}-2\right)}=\frac{\sqrt{x}-1}{-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\frac{1}{\left(\sqrt{x}+1\right)\left(2-\sqrt{x}\right)}\)
\(Q\ge\frac{1}{\frac{\left(\sqrt{x}+1+2-\sqrt{x}\right)^2}{4}}=\frac{4}{3^2}=\frac{4}{9}\)
\(Q_{min}=\frac{4}{9}\) khi \(\sqrt{x}+1=2-\sqrt{x}\Leftrightarrow x=\frac{1}{4}\)
c/ \(R=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}-1}=\sqrt{x}+2+\frac{2}{\sqrt{x}-1}\)
Chắc là bạn ghi nhầm đề, với \(x< 1\) biểu thức này ko có min
Nó chỉ có min khi \(x>1\)
Khi đó: \(R=\sqrt{x}-1+\frac{2}{\sqrt{x}-1}+3\ge2\sqrt{\frac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}}+3=3+2\sqrt{2}\)
\(R_{min}=3+2\sqrt{2}\) khi \(\sqrt{x}-1=\sqrt{2}\Leftrightarrow x=3+2\sqrt{2}\)
