Question

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

Rút gọn biểu thức P = A+B

 

Answer 1

Lời giải:

ĐKXĐ: $x\geq 0; x\neq 9$

\(P=A+B=\frac{2\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}+3)(\sqrt{x}-3)}+\frac{(\sqrt{x}+1)(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}+\frac{11\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}\\ =\frac{(2x-6\sqrt{x})+(x+4\sqrt{x}+3)+(11\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}\\ =\frac{3x+9\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}=\frac{3\sqrt{x}(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}=\frac{3\sqrt{x}}{\sqrt{x}-3}\)