nhầm rồi, để làm lại

a/ \(P=\left[\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right]:\left[\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}}\right]\)

      \(=\left[\frac{4\sqrt{x}\left(2-\sqrt{x}\right)+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right]:\left[\frac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right]\)

        \(=\frac{8\sqrt{x}+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)

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

          \(=\frac{4x}{\sqrt{x}-3}\)

b/ \(P=-1\Rightarrow\frac{4x}{\sqrt{x}-3}=-1\Rightarrow3-\sqrt{x}=4x\Rightarrow4x+\sqrt{x}-3=0\)

                   \(\Rightarrow\orbr{\begin{cases}\sqrt{x}=-1\left(l\right)\\\sqrt{x}=\frac{3}{4}\end{cases}\Rightarrow x=\frac{9}{16}}\)

                                                                 Vậy x = 9/16

ĐKXĐ: x > 0 và \(x\ne4\)

a/ \(P=\left[\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right]:\left[\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}}\right]\)

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

        \(=\frac{8\sqrt{x}-4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{x-\sqrt{x}-2}\)

        \(=\frac{4\sqrt{x}\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

          \(=\frac{4x}{\left(2+\sqrt{x}\right)\left(\sqrt{x}+1\right)}\)

b/ \(P=-1\Rightarrow\frac{4x}{x+3\sqrt{x}+2}=-1\Rightarrow-x-3\sqrt{x}-2=4x\)

                        \(\Rightarrow-5x-3\sqrt{x}-2=0\left(1\right)\), vì (1) > 0 => vô nghiệm

                Vậy k có giá trị nào của x thỏa P = -1

đkxđ: \(x\ge0;x\ne4\)

\(Q=\left[\frac{x-\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}-2}\right]\div\left[\frac{\sqrt{x}+2}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+2}-\frac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\)

\(Q=\left[\frac{x-\sqrt{x}+7+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\div\left[\frac{\left(\sqrt{x}+2\right)^2-\left(\sqrt{x}-2\right)^2-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\)

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

\(Q=\frac{x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{6\sqrt{x}}\)

\(Q=\frac{\left(x+9\right)\sqrt{x}}{6x}\)

\(Q=\frac{x\sqrt{x}+9\sqrt{x}}{6x}\)

đkxđ sửa tí thành \(\hept{\begin{cases}x>0\\x\ne4\end{cases}}\)

\(A=\left[\frac{2\left(x-2\sqrt{x}+1\right)}{x-1}-\frac{2\sqrt{x}-1}{\sqrt{x}+2}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)

\(A=\left[\frac{2\left(x-2\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(x-4\right)\left(\sqrt{x}+2\right)}-\frac{\left(2\sqrt{x}-1\right)\left(x-4\right)}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)

\(A=\left[\frac{2\left(x-2\sqrt{x}+1\right)\left(\sqrt{x}+2\right)-\left(2\sqrt{x}-1\right)\left(x-4\right)}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)

\(A=\left[\frac{x+2\sqrt{x}}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)

\(A=\left[\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(x-4\right)\left(\sqrt{x+2}\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)

\(A=\frac{\sqrt{x}}{x-4}\cdot\frac{\sqrt{x}-2}{\sqrt{x}}\)

\(A=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}\left(x-4\right)}\)

\(A=\frac{\sqrt{x}-2}{x-4}\)

Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé

a, ĐKXĐ: \(x>0;x\ne1;x\ne4\)

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

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

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

\(=\frac{\sqrt{x}-2}{\sqrt{x}}\)

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

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

\(=\dfrac{1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}-2}{3}=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)

b: P=1/4

=>\(\dfrac{\sqrt{x}-2}{3\sqrt{x}}=\dfrac{1}{4}\)

=>\(4\left(\sqrt{x}-2\right)=3\sqrt{x}\)

=>\(4\sqrt{x}-8-3\sqrt{x}=0\)

=>\(\sqrt{x}=8\)

=>x=64

c: Khi \(x=4+2\sqrt{3}\) thì \(P=\dfrac{\sqrt{4+2\sqrt{3}}-2}{3\cdot\sqrt{4+2\sqrt{3}}}\)

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