Question

Answer 1

d: Ta có: \(D=\dfrac{2\sqrt{x}+1}{x-7\sqrt{x}+12}-\dfrac{\sqrt{x}+3}{\sqrt{x}-4}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\)

\(=\dfrac{2\sqrt{x}+1-x+9+2x-8\sqrt{x}+\sqrt{x}-4}{\left(\sqrt{x}-4\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x-5\sqrt{x}+6}{\left(\sqrt{x}-4\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-4}\)

Answer 2

1.

ĐK: \(x\ge0;x\ne16;x\ne9\)

\(D=\dfrac{2\sqrt{x}+1}{x-7\sqrt{x}+12}-\dfrac{\sqrt{x}+3}{\sqrt{x}-4}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)

\(=\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-4\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-4\right)}+\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-4\right)}\)

\(=\dfrac{2\sqrt{x}+1-x+9+2x-7\sqrt{x}-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-4\right)}\)

\(=\dfrac{x-5\sqrt{x}+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-4\right)}\)

\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-4\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}-4}\)

Answer 3

2.

ĐK: \(x\ge0;x\ne1\)

\(E=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}-1}{2}\)

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

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

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

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

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