Question

\(B=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{x+9\sqrt{x}}{9-x};\left(x\ge0;x\ne9;x\ne16\right)\)

\(B=\dfrac{3}{\sqrt{x}-3}+\dfrac{2}{\sqrt{x}+3}+\dfrac{x-5\sqrt{x}-3}{x-9}\)

\(B=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{x+\sqrt{x}}{x-1};\left(x>0;x\ne1\right)\)

Answer 1

1.

\(A=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{x+9\sqrt{x}}{9-x}\)

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

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

\(=\dfrac{2x-6\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x-15\sqrt{x}}{x-9}\)

Answer 2

2.

\(B=\dfrac{3}{\sqrt{x}-3}+\dfrac{2}{\sqrt{x}+3}+\dfrac{x-5\sqrt{x}-3}{x-9}\)

\(=\dfrac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{x-5\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{3\sqrt{x}+9+2\sqrt{x}-6+x-5\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x}{x-9}\)

Answer 3

3.

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

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

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

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

Answer 4

a: Ta có: \(B=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{x+9\sqrt{x}}{9-x}\)

\(=\dfrac{2x-6\sqrt{x}-x-9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x-15\sqrt{x}}{x-9}\)

b: Ta có: \(B=\dfrac{3}{\sqrt{x}-3}+\dfrac{2}{\sqrt{x}+3}+\dfrac{x-5\sqrt{x}-3}{x-9}\)

\(=\dfrac{3\sqrt{x}+9+2\sqrt{x}-6+x-5\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x}{x-9}\)