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Chỉ làm thử thôi nhé-.-
\(B=\left(\sqrt{x+2-4\sqrt{x-2}}+\sqrt{x+2+4\sqrt{x-2}}\right):\sqrt{\frac{4}{x^2}-\frac{4}{x}+1}\left(đk:x\ge2\right)\)
\(=\left(\sqrt{x-2-2\sqrt{x-2}.2+2^2}+\sqrt{x-2+2\sqrt{x-2}.2+2^2}\right):\sqrt{\frac{4}{x^2}-\frac{4x}{x^2}+\frac{x^2}{x^2}}\)
\(=[\left(\sqrt{\left(\sqrt{x-2}-2\right)^2}+\sqrt{\left(\sqrt{x-2}+2\right)^2}\right):\sqrt{\frac{4-4x+x^2}{x^2}}\)
\(=\left(|\sqrt{x-2}-2|+|\sqrt{x-2}+2|\right):\sqrt{\frac{\left(2-x\right)^2}{x^2}}\)
\(=\left(\sqrt{x-2}-2+\sqrt{x-2}+2\right).\frac{x}{2-x}\)
\(=2\sqrt{x-2}.\frac{x}{2-x}=\frac{2x\sqrt{x-2}}{2-x}\)
\(P=\left(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(\frac{2\left(x-2\sqrt{x}+1\right)}{x-1}\right)\)
\(=\left[\frac{\left(x\sqrt{x}-1\right)\left(x+\sqrt{x}\right)}{\left(x-\sqrt{x}\right)\left(x+\sqrt{x}\right)}-\frac{\left(x\sqrt{x}+1\right)\left(x-\sqrt{x}\right)}{\left(x-\sqrt{x}\right)\left(x+\sqrt{x}\right)}\right]:\left[\frac{2\left(\sqrt{x}-1\right)^2}{x-1}\right]\)
Phương trình tương đương :
\(=\frac{2x^2-2x}{x^2-x}:\frac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=2:\frac{2\left(\sqrt{x}-1\right)}{\sqrt{x}+1}=\frac{2\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(B=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right).\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\right)\)
\(=\frac{\left(2\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)}=\frac{2\sqrt{x}+2}{\sqrt{x}+2}\)
\(C=\frac{\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)}\left(\frac{2\sqrt{x}+2}{\sqrt{x}+2}-2\right)=\frac{\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)}.\frac{-2}{\left(\sqrt{x}+2\right)}=\frac{-2}{\sqrt{x}+2}\)
Để C nguyên \(\Rightarrow\sqrt{x}+2=Ư\left(2\right)=2\)
\(\Rightarrow\sqrt{x}=0\Rightarrow x=0\)
a/ ĐKXĐ:...
\(B=\left(\frac{\sqrt{x}+\sqrt{x}+2}{x-4}\right).\frac{x-4}{\sqrt{x}+2}=\frac{2\sqrt{x}+2}{\sqrt{x}+2}\)
\(C=\frac{\sqrt{x}+2}{\sqrt{x}-2}\left(\frac{2\sqrt{x}+2}{\sqrt{x}+2}-2\right)\)
\(C=\frac{\sqrt{x}+2}{\sqrt{x}-2}.\frac{2\sqrt{x}+2-2\sqrt{x}-4}{\sqrt{x}+2}=\frac{-2}{\sqrt{x}-2}\)
Để C đạt GT nguyên
\(\Leftrightarrow\sqrt{x}-2\inƯ_{\left(-2\right)}=\left\{\pm2;\pm1\right\}\)
\(\left[{}\begin{matrix}x=0\\x=9\\x=1\\x=16\end{matrix}\right.\)