a: Sửa đề; \(P=\left(\dfrac{3x+3\sqrt{x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\right)\cdot\left(\dfrac{1}{1-\sqrt{x}}-1\right)\)

\(=\dfrac{3x+3\sqrt{x}-3-x+1+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{1-1+\sqrt{x}}{1-\sqrt{x}}\)

\(=\dfrac{3x+3\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}}{1-\sqrt{x}}=\dfrac{3\sqrt{x}}{1-\sqrt{x}}\)

b: Để \(P=\sqrt{x}\) thì \(3\sqrt{x}=\sqrt{x}-x\)

\(\Leftrightarrow x+2\sqrt{x}=0\)

hay x=0

Lời giải:

a)

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

b)

\(=\frac{(2-\sqrt{x}-\sqrt{x}-3)(2-\sqrt{x}+\sqrt{x}+3)}{1+2\sqrt{x}}=\frac{(-1-2\sqrt{x}).5}{2\sqrt{x}+1}=\frac{-5(2\sqrt{x}+1)}{2\sqrt{x}+1}=-5\)

\(a,\frac{\left(\sqrt{x}+1\right)\cdot\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(x-y\right)\sqrt{x}\left(x+1\right)}\)\(=\frac{\left(\sqrt{x}+1\right)\sqrt{x}\left(x-y\right)}{\left(x-y\right)\sqrt{x} \left(x+1\right)}\)\(=\frac{\sqrt{x}+1}{x+1}\)

\(b,\frac{\left(2-\sqrt{x}\right)^2-\sqrt{x}-3}{1+2\sqrt{x}}=\frac{4+x-4\sqrt{x}-\sqrt{x}-3}{1+2\sqrt{x}}=\frac{1+x-5\sqrt{x}}{1+2\sqrt{x}}\)

a/ ĐKXĐ: \(x\ge0,x\ne1\)

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

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

= \(\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

= \(\dfrac{4\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

= \(\dfrac{4\sqrt{x}}{\sqrt{x}+1}\)

b/ Với \(x\ge0,x\ne1\)

Để \(P=\sqrt{x}-1\Leftrightarrow\dfrac{4\sqrt{x}}{\sqrt{x}+1}=\sqrt{x}-1\)

\(\Leftrightarrow4\sqrt{x}=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)

\(\Leftrightarrow x-4\sqrt{x}-1=0\)

\(\Leftrightarrow\left(\sqrt{x}-2+\sqrt{5}\right)\left(\sqrt{x}-2-\sqrt{5}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-2+\sqrt{5}=0\\\sqrt{x}-2-\sqrt{5}=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2-\sqrt{5}\left(ktm\right)\\\sqrt{x}=2+\sqrt{5}\left(tm\right)\end{matrix}\right.\)

\(\Leftrightarrow x=9+4\sqrt{5}\)

Vậy để \(P=\sqrt{x}-1\) thì \(x=9+4\sqrt{5}\)

\(A=\dfrac{-\left(\sqrt{x}+1\right)\left(2+\sqrt{x}\right)-2\sqrt{x}\left(2-\sqrt{x}\right)+5\sqrt{x}+2}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(\sqrt{x}+2\right)^2}\)

\(A=\dfrac{-3\sqrt{x}-x-2-4\sqrt{x}+2x+5\sqrt{x}+2}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)

\(A=\dfrac{-x-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)

\(A=\dfrac{-\sqrt{x}\left(\sqrt{x}+2\right)^3}{\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)\sqrt{x}\left(3-\sqrt{x}\right)}=\dfrac{-\left(\sqrt{x}+2\right)^2}{\left(2-\sqrt{x}\right)\left(3-\sqrt{x}\right)}\)

CHÚC BẠN HỌC TỐT NHÉ

\(A=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)^2}{-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)\sqrt{x}\left(3-\sqrt{x}\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)

\(=\left(\sqrt{x}+\sqrt{x-1}-\sqrt{x-1}+\sqrt{2}\right)\cdot\left(\dfrac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(2-\sqrt{x}\right)}\right)\)

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