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2. a) \(B=\left(\dfrac{\sqrt{x}+2}{x-2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{2\sqrt{x}}\)

\(B=\left[\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)^2}-\dfrac{\sqrt{x}-2}{x-1}\right]\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{2\sqrt{x}}\)

\(B=\left[\dfrac{\left(\sqrt{x}+2\right)\left(x-1\right)}{\left(\sqrt{x}-1\right)^2\left(x-1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(x-2\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2\left(x-1\right)}\right]\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{2\sqrt{x}}\)

\(B=\left[\dfrac{x\sqrt{x}-\sqrt{x}+2x-2}{\left(\sqrt{x}-1\right)^2\left(x-1\right)}-\dfrac{x\sqrt{x}-4x+5\sqrt{x}-2}{\left(\sqrt{x}-1\right)^2\left(x-1\right)}\right]\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{2\sqrt{x}}\)

\(B=\dfrac{x\sqrt{x}-\sqrt{x}+2x-2-x\sqrt{x}+4x-5\sqrt{x}+2}{\left(\sqrt{x}-1\right)^2\left(x-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{2\sqrt{x}}\)

\(B=\dfrac{6x-6\sqrt{x}}{\left(\sqrt{x}-1\right)^2\left(x-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{2\sqrt{x}}\)

\(B=\dfrac{2\sqrt{x}\left(3\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)^2\left(x-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{2\sqrt{x}}\)

\(B=\dfrac{3\sqrt{x}-3}{x-1}\)

\(B=\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(B=\dfrac{3}{\sqrt{x}+1}\)

b) \(B=1\Leftrightarrow\dfrac{3}{\sqrt{x}+1}=1\)

\(\Leftrightarrow\sqrt{x}+1=3\)

\(\Leftrightarrow\sqrt{x}=2\)

\(\Leftrightarrow x=4\)

\(3x^2-5x+3=0\)

\(\Delta=b^2-4ac\)

   \(=\left(-5\right)^2-4.3.3\)

   \(=-11< 0\)

=> phương trình vô nghiệm

Phương trình hoành độ giao điểm 

x² = 2x - 3 

<=> x² - 2x + 3 = 0

<=> (x - 1)² = - 2 (vô lý)

=> (d) ; (p) không cắt nhau 

=36-12\(\sqrt{5}\)+5+5-2\(\sqrt{35}\)+7=53