Đáp án:a)\(\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
b)\(\frac{3}{13}\)
Giải thích các bước giải:
bài 1:a)M=\(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
=\(\frac{x+2}{(\sqrt{x})^{3}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
=\(\frac{x+2}{(\sqrt{x}-1)(x+\sqrt{x}+1)}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
=\(\frac{x+2+(\sqrt{x}+1)(\sqrt{x}-1)-(x+\sqrt{x}+1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}\)
=\(\frac{x+2+x-1-x-\sqrt{x}-1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}\)
=\(\frac{x-\sqrt{x}}{(\sqrt{x}-1)(x+\sqrt{x}+1)}\)
=\(\frac{\sqrt{x}(\sqrt{x-1})}{(\sqrt{x}-1)(x+\sqrt{x}+1)}\)
=\(\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
b)Khi x=9⇒ \(M=\frac{\sqrt{x}}{x+\sqrt{x}+1}=\frac{\sqrt{9}}{9+\sqrt{9}+1}=\frac{3}{13}\)
