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ĐKXĐ: \(x\ge0\)
\(\frac{1}{\sqrt{x}+1}-\frac{3}{x\sqrt{x}+1}+\frac{2}{x-\sqrt{x}+1}\)
\(=\frac{1}{\sqrt{x}+1}-\frac{3}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}+\frac{2}{x-\sqrt{x}+1}\)
\(=\frac{x-\sqrt{x}+1-3+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}}{x-\sqrt{x}+1}\)
\(=\frac{2\sqrt{x}}{\sqrt{x}-1}-\frac{x+1}{x-1}\)
\(=\frac{2\sqrt{x}\left(x-1\right)-\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x-1\right)}\)
\(=\frac{2x\sqrt{x}-2\sqrt{x}-x\sqrt{x}+x-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x-1\right)}\)
\(=\frac{x\sqrt{x}-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x-1\right)}\)
\(=\frac{x\sqrt{x}-\sqrt{x}+1}{x\sqrt{x}-\sqrt{x}-x+1}=1-\frac{x}{\left(\sqrt{x}-1\right)\left(x-1\right)}\)
a/
\(=\left(\frac{1}{\sqrt{x}+3}+\frac{3}{\sqrt{x}\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+3}-\frac{3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\left(\frac{x-3\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{\sqrt{x}-3}{\sqrt{x}+3}\right)\)
\(=\left(\frac{x-3\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right).\frac{\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\frac{x-3\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)^2}\)
\(=\frac{x-3\sqrt{x}+3}{x\sqrt{x}-6\text{x}+9\sqrt{x}}\)
\(=\frac{x-3\sqrt{x}+3}{x\sqrt{x}-6\text{x}+9\sqrt{x}}\)
b/ Vậy để P>1 khi BT trên>1
Ta có phương trình tương đương
\(x-3\sqrt{x}+3-x\sqrt{x}+6\text{x}-9>0\)
\(-x\sqrt{x}+7\text{x}-3\sqrt{x}-6>0\)
Giải pt rồi suy ra
tick cho mình nha
a)\(P=\left(\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right):\left(\frac{\sqrt{x}}{\sqrt{x}-1}-1\right)ĐK:\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}.\)
\(=\left(\frac{\sqrt{x}+1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\left(\frac{\sqrt{x}-1}{\sqrt{x}-\sqrt{x}+1}\right)\)
=\(\frac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2\sqrt{x}+1}{\sqrt{x}+1}\)
b)P=3/2 <=>\(\frac{2\sqrt{x}+1}{\sqrt{x}+1}=\frac{3}{2}\Leftrightarrow2\sqrt{x}+1=\frac{3}{2}\sqrt{x}+\frac{3}{2}.\)
\(\Leftrightarrow\frac{1}{2}\sqrt{x}=\frac{1}{2}\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)
Với x=1 thoả nãm yêu cầu
\(S=\frac{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)+\left(2x-\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-3x\sqrt{x}+2x-\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(S=\frac{x\sqrt{x}-2x+2\sqrt{x}-1+2x\sqrt{x}+x-2\sqrt{x}-1-3x\sqrt{x}+2x-\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(S=\frac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(S=\frac{1}{\sqrt{x}+1}\)
Vậy \(S=\frac{1}{\sqrt{x}+1}\)
\(\frac{1}{\sqrt{x-1}-\sqrt{x}}+\frac{1}{\sqrt{x-1}+\sqrt{x}}+\frac{\sqrt{x^3}-x}{\sqrt{x}-1}\)
\(ĐKXĐ:x\ge1\)
\(\frac{1}{\sqrt{x-1}-\sqrt{x}}+\frac{1}{\sqrt{x-1}+\sqrt{x}}+\frac{x\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(\frac{\sqrt{x-1}+\sqrt{x}+\sqrt{x-1}-\sqrt{x}}{x-1-x}+x\)
\(\frac{2\sqrt{x-1}}{-1}+x\)
\(x-2\sqrt{x-1}\)
\(\left(\sqrt{x-1}-1\right)^2\)