Chỉ làm thử thôi nhé-.-

\(B=\left(\sqrt{x+2-4\sqrt{x-2}}+\sqrt{x+2+4\sqrt{x-2}}\right):\sqrt{\frac{4}{x^2}-\frac{4}{x}+1}\left(đk:x\ge2\right)\)

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

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

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

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

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

1) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)

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

\(\Leftrightarrow P=\frac{\left(2+\sqrt{x}\right)^2-\left(2-\sqrt{x}\right)^2+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)

\(\Leftrightarrow P=\frac{4+4\sqrt{x}+x-4+4\sqrt{x}-x+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)

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

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

2) Để \(P=2\)

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

\(\Leftrightarrow4\sqrt{x}=4-2\sqrt{x}\)

\(\Leftrightarrow6\sqrt{x}=4\)

\(\Leftrightarrow\sqrt{x}=\frac{2}{3}\)

\(\Leftrightarrow x=\frac{4}{9}\)

Vậy để \(P=2\Leftrightarrow x=\frac{4}{9}\)

3) Khi \(\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-2=0\\2\sqrt{x}-1==0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=2\\\sqrt{x}=\frac{1}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\left(ktm\right)\\x=\frac{1}{4}\left(tm\right)\end{cases}}\)

Thay \(x=\frac{1}{4}\)vào P, ta được :

\(\Leftrightarrow P=\frac{4\sqrt{\frac{1}{4}}}{2-\sqrt{\frac{1}{4}}}=\frac{4\cdot\frac{1}{2}}{2-\frac{1}{2}}=\frac{2}{\frac{3}{2}}=\frac{4}{3}\)

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

\(\Leftrightarrow\frac{4\sqrt{x}}{2-\sqrt{x}}=\frac{\sqrt{x}+3}{2\sqrt{x}-1}\)

\(\Leftrightarrow8x-4\sqrt{x}=-x-\sqrt{x}+6\)

\(\Leftrightarrow9x-3\sqrt{x}-6=0\)

\(\Leftrightarrow3x-\sqrt{x}-2=0\)

\(\Leftrightarrow\sqrt{x}=3x-2\)

\(\Leftrightarrow x=9x^2-12x+4\)

\(\Leftrightarrow9x^2-13x+4=0\)

\(\Leftrightarrow\left(9x-4\right)\left(x-1\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}9x-4=0\\x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{9}\\x=1\end{cases}}\)

Thử lại ta được kết quá : \(x=\frac{4}{9}\left(ktm\right)\); \(x=1\left(tm\right)\)

Vậy để \(P=\frac{\sqrt{x}+3}{2\sqrt{x}-1}\Leftrightarrow x=1\)

5) Để biểu thức nhận giá trị nguyên

\(\Leftrightarrow\frac{4\sqrt{x}}{2-\sqrt{x}}\inℤ\)

\(\Leftrightarrow4\sqrt{x}⋮2-\sqrt{x}\)

\(\Leftrightarrow-4\left(2-\sqrt{x}\right)+8⋮2-\sqrt{x}\)

\(\Leftrightarrow8⋮2-\sqrt{x}\)

\(\Leftrightarrow2-\sqrt{x}\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{1;3;0;4;-2;6;-6;10\right\}\)

Ta loại các giá trị < 0

\(\Leftrightarrow\sqrt{x}\in\left\{1;3;0;4;6;10\right\}\)

\(\Leftrightarrow x\in\left\{1;9;0;16;36;100\right\}\)

Vậy để \(P\inℤ\Leftrightarrow x\in\left\{1;9;0;16;36;100\right\}\)

\(\)

a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne1\\x\ne16\end{cases}}\)

\(B=\frac{2\left(x+4\right)}{x-3\sqrt{x}-4}+\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{8}{\sqrt{x}-4}\)

\(\Leftrightarrow B=\frac{2x+8+\sqrt{x}\left(\sqrt{x}-4\right)-8\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)

\(\Leftrightarrow B=\frac{2x+8+x-4\sqrt{x}-8\sqrt{x}-8}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)

\(\Leftrightarrow B=\frac{3x-12\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)

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

b) Để B nguyên'

\(\Leftrightarrow3\sqrt{x}⋮\sqrt{x}+1\)

\(\Leftrightarrow3\left(\sqrt{x}+1\right)-3⋮\sqrt{x}+1\)

\(\Leftrightarrow3⋮\sqrt{x}+1\)

\(\Leftrightarrow\sqrt{x}+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{0;2\right\}\)(Đã loại những giá trị âm)

\(\Leftrightarrow x\in\left\{0;4\right\}\)

Vậy để \(B\inℤ\Leftrightarrow x\in\left\{0;2\right\}\)

Loại giá trị 0 ở câu b cho mik nhé (Vì ktm đkxđ)