Giải câu 6 trang 28 toán VNEN 9 tập 1.
a) Ta có:
M = $\frac{x\sqrt{x} - 1}{x - \sqrt{x}}$ - $\frac{x\sqrt{x} + 1}{x + \sqrt{x}}$ + $\frac{x + 1}{\sqrt{x}}$
= $\frac{(\sqrt{x})^{3} - 1}{\sqrt{x}(\sqrt{x} - 1)}$ - $\frac{(\sqrt{x})^{3} + 1}{\sqrt{x}(\sqrt{x} + 1)}$ + $\frac{x + 1}{\sqrt{x}}$
= $\frac{(\sqrt{x} - 1)((\sqrt{x})^{2} + \sqrt{x} + 1)}{\sqrt{x}(\sqrt{x} - 1)}$ - $\frac{(\sqrt{x} + 1)((\sqrt{x})^{2} - \sqrt{x} + 1)}{\sqrt{x}(\sqrt{x} + 1)}$ + $\frac{x + 1}{\sqrt{x}}$
= $\frac{(\sqrt{x})^{2} + \sqrt{x} + 1}{\sqrt{x}}$ - $\frac{(\sqrt{x})^{2} - \sqrt{x} + 1}{\sqrt{x}}$ + $\frac{x + 1}{\sqrt{x}}$
= $\frac{(\sqrt{x})^{2} + \sqrt{x} + 1 - (\sqrt{x})^{2} + \sqrt{x} - 1}{\sqrt{x}}$ + $\frac{x + 1}{\sqrt{x}}$
= $\frac{2\sqrt{x}}{\sqrt{x}}$ + $\frac{x + 1}{\sqrt{x}}$
= 2 + $\frac{x + 1}{\sqrt{x}}$
b) M = $\frac{9}{2}$
$\Leftrightarrow $ 2 + $\frac{x + 1}{\sqrt{x}}$ = $\frac{9}{2}$
$\Leftrightarrow $ $\frac{x + 1}{\sqrt{x}}$ = $\frac{5}{2}$
$\Leftrightarrow $ x + 1 = $\frac{5}{2}$$\sqrt{x}$
$\Leftrightarrow $ $\sqrt{x}$ = 2 hoặc $\sqrt{x}$ = $\frac{1}{2}$
$\Leftrightarrow $ x = 4 hoặc x = $\frac{1}{4}$
Vậy S = {4 ; $\frac{1}{4}$}.
c) M = 2 + $\frac{x + 1}{\sqrt{x}}$ = 2 + $\sqrt{x}$ + $\frac{1}{\sqrt{x}}$
Áp dụng bất đẳng thức cô -si ta có: $\sqrt{x}$ + $\frac{1}{\sqrt{x}}$ $\geq $ 2.$\sqrt{\sqrt{x}.\frac{1}{\sqrt{x}}}$ = 2
Suy ra: M $\geq $ 2 + 2 = 4.
Vậy M $\geq $ 4.