Toán 9 Tìm giá trị nhỏ nhất, lớn nhất

f1/ P=[TEX]x-1 \geq -1[/TEX]
f2/ Q=[TEX]-\frac{1}{(\sqrt{x}-2)(\sqrt{x}+1)}=\frac{1}{(2-\sqrt{x})(\sqrt{x}+1)} \geq \frac{1}{(\frac{2-\sqrt{x}+\sqrt{x}+1}{2})^2}=\frac{4}{9}[/TEX]
f3/[tex]R=\frac{\sqrt{x}(\sqrt{x}+1)}{\sqrt{x}-1}=\sqrt{x}+2+\frac{2}{\sqrt{x}-1}=\sqrt{x}-1+\frac{2}{\sqrt{x}-1}+3\geq 2\sqrt{2}+3[/tex]
i1/[tex]B=\frac{\sqrt{x}+3}{\sqrt{x}+1}=1+\frac{2}{\sqrt{x}+1}\leq 3[/tex]
i2/ [tex]\frac{1}{C}=\frac{x+8\sqrt{x}+7}{\sqrt{x}-1}=\sqrt{x}+9+\frac{16}{\sqrt{x}-1}=\sqrt{x}-1+\frac{16}{\sqrt{x}-1}+10\geq 8+10=18[/tex]