làm hộ tui c5 ahhjjj

Đáp án:

a.Đúng

b.Đúng

c.Đúng

d.Sai 

Giải thích các bước giải:

Ta có:

$\begin{align*}
P &= \left( \dfrac{x+9}{x-9} - \dfrac{\sqrt{x}}{\sqrt{x}+3} \right) : \dfrac{2\sqrt{x}+2}{x-3\sqrt{x}} \\
&= \left( \dfrac{x+9}{(\sqrt{x}-3)(\sqrt{x}+3)} - \dfrac{\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)} \right) : \dfrac{2(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-3)} \\
&= \left( \dfrac{x+9 - (x-3\sqrt{x})}{(\sqrt{x}-3)(\sqrt{x}+3)} \right) \cdot \dfrac{\sqrt{x}(\sqrt{x}-3)}{2(\sqrt{x}+1)} \\
&= \dfrac{3\sqrt{x}+9}{(\sqrt{x}-3)(\sqrt{x}+3)} \cdot \dfrac{\sqrt{x}(\sqrt{x}-3)}{2(\sqrt{x}+1)} \\
&= \dfrac{3(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)} \cdot \dfrac{\sqrt{x}(\sqrt{x}-3)}{2(\sqrt{x}+1)} \\
&= \dfrac{3}{\sqrt{x}-3} \cdot \dfrac{\sqrt{x}(\sqrt{x}-3)}{2(\sqrt{x}+1)} \\
&= \dfrac{3\sqrt{x}}{2(\sqrt{x}+1)} \\
&= \dfrac{3\sqrt{x}}{2\sqrt{x}+2}
\end{align*}$

a.Ta có:

$x = 4 - 2\sqrt{3} = 3 - 2\sqrt{3} + 1 = (\sqrt{3}-1)^2$

$\to \sqrt{x} = \sqrt{(\sqrt{3}-1)^2} = |\sqrt{3}-1| = \sqrt{3}-1$

$\to P = \dfrac{3(\sqrt{3}-1)}{2(\sqrt{3}-1) + 2}$

$\to P = \dfrac{3\sqrt{3}-3}{2\sqrt{3}}$

$\to P = \dfrac{\sqrt{3}(3\sqrt{3}-3)}{2\sqrt{3}\cdot\sqrt{3}} = \dfrac{9-3\sqrt{3}}{6} = \dfrac{3(3-\sqrt{3})}{6} = \dfrac{3-\sqrt{3}}{2}$

d.Ta có:

$P<1$

$\to \dfrac{3\sqrt{x}}{2\sqrt{x}+2} < 1$

$\to 3\sqrt{x} < 2\sqrt{x} + 2$

$\to \sqrt{x}<2$

$\to 0\le x<4$

Mà $x>0, x\ne 9$

$\to 0<x<4$

$\to x\in\{1, 2, 3\}$ vì $x\in Z$