Rút gọn biểu thức : (2√x / x – √x – 6 + √x / √x – 3 ) ÷ √x / √x – 3 07/11/2021 Bởi Katherine Rút gọn biểu thức : (2√x / x – √x – 6 + √x / √x – 3 ) ÷ √x / √x – 3
$\Bigg(\dfrac{2\sqrt{x}}{x-\sqrt{x}-6}+\dfrac{\sqrt{x}}{\sqrt{x}-3}\Bigg):\dfrac{\sqrt{x}}{\sqrt{x}-3}\,\,(x>0,\,x\ne9)\\=\Bigg[\dfrac{2\sqrt{x}}{(\sqrt{x}+2)(\sqrt{x}-3)}+\dfrac{\sqrt{x}}{\sqrt{x}-3}\Bigg].\dfrac{\sqrt{x}-3}{\sqrt{x}}\\=\dfrac{2\sqrt{x}+\sqrt{x}(\sqrt{x}+2)}{(\sqrt{x}+2)(\sqrt{x}-3)}.\dfrac{\sqrt{x}-3}{\sqrt{x}}\\=\dfrac{2\sqrt{x}+x+2\sqrt{x}}{\sqrt{x}(\sqrt{x}+2)}\\=\dfrac{x+4\sqrt{x}}{\sqrt{x}(\sqrt{x}+2)}\\=\dfrac{\sqrt{x}(\sqrt{x}+4)}{\sqrt{x}(\sqrt{x}+2)}\\=\dfrac{\sqrt{x}+4}{\sqrt{x}+2}$ Bình luận
$\Bigg(\dfrac{2\sqrt{x}}{x-\sqrt{x}-6}+\dfrac{\sqrt{x}}{\sqrt{x}-3}\Bigg):\dfrac{\sqrt{x}}{\sqrt{x}-3}\,\,(x>0,\,x\ne9)\\=\Bigg[\dfrac{2\sqrt{x}}{(\sqrt{x}+2)(\sqrt{x}-3)}+\dfrac{\sqrt{x}}{\sqrt{x}-3}\Bigg].\dfrac{\sqrt{x}-3}{\sqrt{x}}\\=\dfrac{2\sqrt{x}+\sqrt{x}(\sqrt{x}+2)}{(\sqrt{x}+2)(\sqrt{x}-3)}.\dfrac{\sqrt{x}-3}{\sqrt{x}}\\=\dfrac{2\sqrt{x}+x+2\sqrt{x}}{\sqrt{x}(\sqrt{x}+2)}\\=\dfrac{x+4\sqrt{x}}{\sqrt{x}(\sqrt{x}+2)}\\=\dfrac{\sqrt{x}(\sqrt{x}+4)}{\sqrt{x}(\sqrt{x}+2)}\\=\dfrac{\sqrt{x}+4}{\sqrt{x}+2}$