cho C= $\frac{1}{căn x+1}$ -$\frac{3}{x.căn x+1}$ +$\frac{2}{x-căn x+1}$ a)rút gọn biểu thức C b)chứng minh C<1

Giải thích các bước giải:

Ta có:

$C=\dfrac{1}{\sqrt{x}+1}-\dfrac{3}{x\sqrt{x}+1}+\dfrac{2}{x-\sqrt{x}+1}$

$\to C=\dfrac{x-\sqrt{x}+1}{(\sqrt{x}+1)(x-\sqrt{x}+1)}-\dfrac3{(\sqrt{x}+1)(x-\sqrt{x}+1)}+\dfrac{2(\sqrt{x}+1)}{(\sqrt{x}+1)(x-\sqrt{x}+1)}$

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

$\to C=\dfrac{x+\sqrt{x}}{(\sqrt{x}+1)(x-\sqrt{x}+1)}$

$\to C=\dfrac{\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}+1)(x-\sqrt{x}+1)}$

$\to C=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}$

b.Ta có:

$C=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}$

$\to 1-C=1-\dfrac{\sqrt{x}}{x-\sqrt{x}+1}$

$\to 1-C=\dfrac{x-\sqrt{x}+1-\sqrt{x}}{x-\sqrt{x}+1}$

$\to 1-C=\dfrac{x-2\sqrt{x}+1}{x-\sqrt{x}+1}$

$\to 1-C=\dfrac{(\sqrt{x}-1)^2}{x-\sqrt{x}+1}$

Mà $x-\sqrt{x}+1=(\sqrt{x}-\dfrac12)^2+\dfrac34>0$

$\to \dfrac{(\sqrt{x}-1)^2}{x-\sqrt{x}+1}\ge 0$

$\to 1-C\ge 0$

$\to C\le 1$