cho biểu thức A =[tex]\bg_white \left ( \frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1} \right )[/tex] x [tex]\bg_white \frac{x^{2}-2x+1}{2}[/tex]
a) Tìm ĐKXĐ
b) Rút gọn A
c) CMR : 0<x<1 thì A>0
d) Tính A khi x = [tex]\bg_white 4+ 2\sqrt{3}[/tex]
e tìm GTNN của A
a) ĐKXĐ: $x\geq 0;x\neq 1$
b) $A=(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}).\dfrac{x^2-2x+1}{2}
\\=\left [ \dfrac{\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+1)}-\dfrac{\sqrt{x}+2}{(\sqrt{x}+1)^2} \right ].\dfrac{(x-1)^2}{2}
\\=\dfrac{(\sqrt x-2)(\sqrt{x}+1)-(\sqrt{x}+2)(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)^2}.\dfrac{(
\sqrt x+1)^2(\sqrt x-1)^2}{2}
\\=\dfrac{(x-\sqrt x-2-x-\sqrt x+2)(\sqrt x-1)}{2}
\\=\dfrac{-2\sqrt{x}(\sqrt{x}-1)}{2}=\sqrt x-x$
c) $0<x<1\Rightarrow \sqrt{x}<1\Leftrightarrow \sqrt{x}-1<0\Leftrightarrow -\sqrt{x}(\sqrt{x}-1)>0$ hay $A>0$
d) $A=\sqrt{4+2\sqrt{3}}-(4+2\sqrt{3})=\sqrt{(\sqrt{3}+1)^2}-4-2\sqrt{3}=\sqrt{3}+1-4-2\sqrt{3}=-3-\sqrt{3}$
e) GTLN chứ nhỉ?