Đáp án:
a.$N=x-\sqrt{x}+1$
b.$N\ge\dfrac34$
c.$x\in\{0, 1, \dfrac{7\pm3\sqrt{5}}{2}\}$
Giải thích các bước giải:
a.Ta có:
$N=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2(x-1)}{\sqrt{x}-1}$
$\to N=\dfrac{\sqrt{x}((\sqrt{x})^3-1)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}(2\sqrt{x}+1)}{\sqrt{x}}+\dfrac{2(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}-1}$
$\to N=\dfrac{\sqrt{x}(\sqrt{x}-1)(x+\sqrt{x}+1)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}(2\sqrt{x}+1)}{\sqrt{x}}+\dfrac{2(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}-1}$
$\to N=\sqrt{x}(\sqrt{x}-1)-(2\sqrt{x}+1)+2(\sqrt{x}+1)$
$\to N=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2$
$\to N=x-\sqrt{x}+1$
b.Ta có:
$N=x-\sqrt{x}+1$
$\to N=(\sqrt{x}-\dfrac12)^2+\dfrac34\ge\dfrac34$
$\to GTNN_N=\dfrac34$ khi đó $\sqrt{x}-\dfrac12=0\to x=\dfrac14$
c.Ta có:
$M=\dfrac{2\sqrt{x}}N$
$\to M=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}$
Từ câu b $\to N>0$ do $x\ge 0\to M\ge 0$
Mà $M=\dfrac{2}{\sqrt{x}-1+\dfrac1{\sqrt{x}}}$
$\to M=\dfrac{2}{\sqrt{x}+\dfrac1{\sqrt{x}}-1}$
$\to M\le \dfrac{2}{2\sqrt{\sqrt{x}\cdot \dfrac1{\sqrt{x}}}-1}$
$\to M\le 2$
$\to 0\le M\le 2$
Do $M\in Z\to M\in\{0,1,2\}$
Nếu $M=0\to \dfrac{2\sqrt{x}}{x-\sqrt{x}+1}=0\to x=0$
Nếu $M=1\to \dfrac{2\sqrt{x}}{x-\sqrt{x}+1}=1\to x=\dfrac{7\pm3\sqrt{5}}{2}$
Nếu $M=2\to \dfrac{2\sqrt{x}}{x-\sqrt{x}+1}=2\to x=1$
$\to x\in\{0, 1, \dfrac{7\pm3\sqrt{5}}{2}\}$
