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ĐKXĐ: \(x>0;x\ne1\)
\(A=\left(\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)+\left(\frac{x-1}{\sqrt{x}}\right)\left(\frac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(A=\left(\frac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\right)+\frac{2x+2}{\sqrt{x}}\)
\(A=2+\frac{2x+2}{\sqrt{x}}=\frac{2\left(x+\sqrt{x}+1\right)}{\sqrt{x}}\)
\(A=2\left(\sqrt{x}+\frac{1}{\sqrt{x}}+1\right)\ge2\left(2\sqrt{\sqrt{x}.\frac{1}{\sqrt{x}}}+1\right)=6\)
Dấu "=" xảy ra khi \(x=1\) ko phù hợp ĐKXĐ nên \(A_{min}\) ko tồn tại
ĐKXĐ:\(x\ne1;x>0\)
\(A=\left(\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)+\frac{x-1}{\sqrt{x}}.\frac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2}{x-1}\)
\(A=\frac{x+\sqrt{x}+1-x+\sqrt{x}-1+x+2\sqrt{x}+1+x-2\sqrt{x}+1}{\sqrt{x}}\)
\(A=\frac{2x+2\sqrt{x}+2}{\sqrt{x}}=2\sqrt{x}+2+\frac{2}{\sqrt{x}}\ge2\sqrt{2\sqrt{x}.\frac{2}{\sqrt{x}}}+2=6\)
"="\(\Leftrightarrow x=1\)
ĐKXĐ: \(x\ge4\)
a/ \(A=\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\left[\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right]\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\left(\frac{x-4-x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(-3\right)}\)
\(=\frac{\sqrt{x}-2}{-3\sqrt{x}}\)
b/ A = 0 \(\Rightarrow\frac{\sqrt{x}-2}{-3\sqrt{x}}=0\Rightarrow\sqrt{x}-2=0\Rightarrow\sqrt{x}=2\Rightarrow x=4\)
\(A=\left(\frac{\sqrt{x}}{x-1}-\frac{1}{\sqrt{x}+1}\right):\frac{\sqrt{x}-1}{x-1}\)
\(=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\frac{\sqrt{x}-1}{x-1}\)
\(=\frac{1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{x-1}{\sqrt{x}-1}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}=\frac{1}{\sqrt{x}-1}\)