Câu II:
1: ĐKXĐ: x>=0; x<>1
\(C=\left(\dfrac{2\sqrt{x}}{\sqrt{x}-1}-\dfrac{x+1}{x-2\sqrt{x}+1}:\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\cdot\dfrac{\sqrt{x}}{x\sqrt{x}+1}\)
\(=\left(\dfrac{2\sqrt{x}}{\sqrt{x}-1}-\dfrac{x+1}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}\right)\cdot\dfrac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\left(\dfrac{2\sqrt{x}}{\sqrt{x}-1}-\dfrac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{2x-x-1}{\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{x-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}}{x-\sqrt{x}+1}=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\)
2: ĐKXĐ: \(x\in R\)
\(D=\sqrt{x^2-2x+1}+\sqrt{x^2-6x+9}\)
\(=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-3\right)^2}\)
\(=\left|x-1\right|+\left|x-3\right|=\left|x-1\right|+\left|3-x\right|>=\left|x-1+3-x\right|=2\)
Dấu '=' xảy ra khi (x-1)(x-3)<=0
=>1<=x<=3
