\(P=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)\(\left(ĐKXĐ:x\ne4\right)\)

\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{-2-5\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

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

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

\(P=\frac{3\sqrt{x}}{\sqrt{x}+2}\)

b) Với  \(x=3\)( thỏa mãn ĐKXĐ ) ta có  \(P=\frac{3\sqrt{3}}{\sqrt{3}+2}=-9+6\sqrt{3}\)

c) A ở đâu ???? '-' 

a) Ta có:

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

\(A=\frac{\sqrt{4}-3}{4-\sqrt{4}+1}\)

\(A=\frac{2-3}{4-2+1}=-\frac{1}{3}\)

b) đk: \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)

\(B=\left(\frac{3\sqrt{x}+6}{x-9}-\frac{2}{\sqrt{x}-3}\right):\frac{1}{\sqrt{x}+3}\)

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

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

\(B=\frac{\sqrt{x}}{\sqrt{x}-3}\)

c) \(P=AB\)

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

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

Vì \(\left|P\right|=P\Leftrightarrow\orbr{\begin{cases}P=P\\P=-P\end{cases}}\Leftrightarrow\orbr{\begin{cases}0=0\left(tm\right)\\\sqrt{x}=-\sqrt{x}\end{cases}}\Rightarrow\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)

a/ Vì \(x\ge3>-\frac{2}{3}\) nên giá trị biểu thức là : 

\(x+\frac{2}{3}+x-3=2x-\frac{7}{3}\)

b/ Vì \(x>2>\frac{4}{3}>-\frac{2}{5}\) nên giá trị biểu thức là : 

\(-\left(x+\frac{2}{5}\right)+\left(x-\frac{4}{3}\right)=-\frac{4}{3}-\frac{2}{5}=-\frac{26}{15}\)

1. P = \(\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)                       ĐKXĐ: \(x\ne-3\),  \(x\ne2\)

       = \(\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{1}{x-2}\)

       = \(\frac{x^2-4}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{x+3}{x-2}\)

       = \(\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)

       = \(\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)

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

       = \(\frac{x-4}{x-2}\)

2. P=\(\frac{-3}{4}\)

<=> \(\frac{x-4}{x-2}=\frac{-3}{4}\)

<=> 4 ( x - 4 ) = -3  ( x - 2 )

<=> 4x - 16 = -3x + 6

<=> 7x = 2 

<=> x = \(\frac{22}{7}\)

3. \(x^2-9=0\)

<=> ( x -3 ) ( x + 3 ) = 0

<=> \(\orbr{\begin{cases}x=3\left(tm\right)\\x=-3\left(ktm\right)\end{cases}}\)

-> P = \(\frac{3-4}{3-2}\) = -1

A = |2x - 5| + 3 - 2x

A = 2x - 5 + 3 - 2x

A = (2x - 2x) + (-5 + 3)

A = -2

B = |x² - 5x + 4| - 4 + 5x - x²

B = x² - 5x + 4 - 4 + 5x - x²

B = (x² - x²) + (-5x + 5x) + (4 - 4)

B = 0