\(C=\left(\frac{2\sqrt{x}}{\sqrt{x}-3}+\frac{\sqrt{x}}{\sqrt{x}-3}+\frac{3x+3}{9-x}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}-3}-\frac{1}{2}\right)\) ĐK \(x\ge0;x\ne9\)

\(C=\left(\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x+3}\right)}-\frac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\frac{2\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-3\right)}-\frac{1\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}-3\right)}\right)\)

\(C=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{2\left(\sqrt{x}-3\right)}\right)\)

\(C=\frac{-3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\frac{\sqrt{x}+1}{2\left(\sqrt{x}-3\right)}\)

\(C=\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\) x \(\frac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}\)

\(C=\frac{-6}{\sqrt{x}+3}\)

b: ta có \(C=\frac{-6}{\sqrt{x}+3}\) mà \(C=\frac{1}{2}\)

\(\frac{-6}{\sqrt{x}+3}=\frac{1}{2}\)

\(-12=\sqrt{x}+3\)

\(\sqrt{x}=-15\)(Loại)

=> x không có giá trị nào để C=\(\frac{1}{2}\)

đề không sai đâu bạn

1)P=\(\left(\frac{2\sqrt{x}\cdot\left(\sqrt{x}-3\right)}{\sqrt{x}+3}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\sqrt{x}-3}\right):\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)

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

2) P <\(\frac{-1}{2}\)\(\Rightarrow-\frac{3}{\sqrt{x}+1}< -\frac{1}{2}\)\(\Leftrightarrow\sqrt{x}< 5\)\(\Leftrightarrow x< 25\)

vậy: khi \(0\le x< 25\)và \(x\ne9\)thì P<\(\frac{-1}{2}\)

Bài 1:

a)  \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)

\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)

\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)

\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)

b)   \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)

\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)

\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)

\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)

c)  ĐK:  \(a\ge0;a\ne1\)

  \(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)

\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)

\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)

\(=1-a+a=1\)

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

\(\frac{3.\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3}{\sqrt{x}+3}\)

từ dòng cuối là sai rồi bạn à

Bạn bỏ dòng cuối đi còn lại đúng rồi

Ở tử đặt nhân tử chung căn x chung  rồi lại đặt căn x +1 chung

Ở mẫu tách 3 căn x ra 2 căn x +căn x rồi đặt nhân tử 2 căn x ra 

rút gọn được \(\frac{3\sqrt{x}-5}{2\sqrt{x}+1}\)

cảm ơn bạn nha