M=(\(\frac{\sqrt{x}}{\sqrt{x}+1}\)-1): \(\frac{-1}{x+\sqrt{x}+1}\)

M=\(\frac{-1}{\sqrt{x}+1}\).  -(x+\(\sqrt{x}\)+1)

M=\(\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\)

b, x=1 

M = \(\frac{3}{2}\)

c, M= 0 

=> x +\(\sqrt{x}\)+1= 0

mặt khác x+\(\sqrt{x}\)+1 = (\(\sqrt{x}\)+0,5)²+0,75 >0

=> x vô nghiệm........

a) \(C=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)

\(C=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)

\(C=\frac{3\sqrt{x}-x+x+9}{9-x}:\frac{3\sqrt{x}+1-\sqrt{x}+3}{x-3\sqrt{x}}\)

\(C=\frac{3\sqrt{x}+9}{9-x}:\frac{2\sqrt{x}+4}{x-3\sqrt{x}}\)

\(C=\frac{3\left(\sqrt{x}+3\right)\cdot\sqrt{x}\left(\sqrt{x}-3\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)\cdot2\left(\sqrt{x}+2\right)}\)

\(C=\frac{3\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)

b) Dễ thấy \(C=\frac{3\sqrt{x}}{2\left(\sqrt{x}+2\right)}\ge0\forall x\)do đó không có giá trị của x thỏa mãn \(C< -1\)

phần b hình như phải xét hiệu nhỉ.

a, \(\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)

\(=\left(\frac{-3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):(\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}+3\right)})\)

\(=\frac{-3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{2\sqrt{x}+4}\)

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

câu b mình quên cách tính rồi sorry