Question

Cho A= \(\dfrac{2\sqrt{x}}{\sqrt{x}+3}\)và B = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{x-9}\)

Thu gọn B và tính M=A+B

Answer 1

\(B=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\left(x\ne\sqrt{3},x\ge0\right)\)

\(B=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(B=\dfrac{x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(B=\dfrac{x+14\sqrt{x}}{x-9}\)

\(M=A+B=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{x+14\sqrt{x}}{x-9}\)

\(M=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+x+14\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{2x-6\sqrt{x}+x+14\sqrt{x}}{x-9}\)

\(M=\dfrac{3x+8\sqrt{x}}{x-9}\)