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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
\(A=\frac{1}{\sqrt{x-1}-\sqrt{x}}+\frac{1}{\sqrt{x-1}+\sqrt{x}}+\frac{\sqrt{x^3}-x}{\sqrt{x}-1}\) \(ĐKXĐ:x\ne\pm1\)
\(=\frac{\sqrt{x-1}+\sqrt{x}+\sqrt{x-1}-\sqrt{x}}{\left(\sqrt{x-1}-\sqrt{x}\right)\left(\sqrt{x-1}+\sqrt{x}\right)}+\frac{x\sqrt{x}-x}{\sqrt{x}-1}\)
\(=\frac{2\sqrt{x-1}}{x-1-x}+\frac{x\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(=x-2\sqrt{x-1}\)
Câu c mình ko làm được
a)
\(A=\left(\frac{1}{1-\sqrt{3}}-\frac{1}{1+\sqrt{3}}\right):\frac{1}{\sqrt{3}}\\ =\left(\frac{1+\sqrt{3}-1+\sqrt{3}}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}\right):\frac{1}{\sqrt{3}}\\ =\frac{2\sqrt{3}}{1-\left(\sqrt{3}\right)^2}:\frac{1}{\sqrt{3}}\\ =\frac{2\sqrt{3}}{-2}\cdot\sqrt{3}=-3\)
\(B=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2\sqrt{x}-1}{x-\sqrt{x}}\left(ĐK:x>0;x\ne1\right)\\ =\frac{\left(\sqrt{x}\right)^2}{\sqrt{x}\left(\sqrt{x-1}\right)}-\frac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\frac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}}\)
b)
\(A=\frac{1}{6}B\Leftrightarrow-3=\frac{1}{6}\cdot\frac{\sqrt{x}-1}{\sqrt{x}}\\ \Leftrightarrow-18=\frac{\sqrt{x}-1}{\sqrt{x}}\Leftrightarrow-18\sqrt{x}=\sqrt{x}-1\\ \Leftrightarrow-19\sqrt{x}=-1\\ \Leftrightarrow\sqrt{x}=\frac{1}{19}\Rightarrow x=\frac{1}{361}\)
Vậy với x = \(\frac{1}{361}\)thì \(A=\frac{1}{6}B\)
Có gì sai mọi người góp ý nha!
\(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}}\)
\(=-1+\frac{1}{2\sqrt{x}-1}+\frac{x\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(=\frac{-2\sqrt{x}+1+1+2\text{x}\sqrt{x}-x}{2\sqrt{x}-1}\)
\(=\frac{-2\sqrt{x}+2+2\text{x}\sqrt{x}-x}{2\sqrt{x}-1}\)
tick cho mình nha