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\(a\sqrt{a}+2a+\sqrt{a}+2=\left(a\sqrt{a}+2a\right)+\left(\sqrt{a}+2\right)\)
\(=a\left(\sqrt{a}+2\right)+\left(\sqrt{a}+2\right)=\left(\sqrt{a}+2\right)\left(a+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(5\sqrt{14}-\sqrt{21}=\sqrt{7}\left(5\sqrt{2}-\sqrt{3}\right)\)
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\(=\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\)
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\(\sqrt{ab}-\sqrt{a}-\sqrt{b}+1\)
\(=\sqrt{a}\left(\sqrt{b}-1\right)-\left(\sqrt{b}-1\right)\)
\(=\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\)
\(\sqrt{ab}-\sqrt{a}-\sqrt{b}+1=\sqrt{a}\left(\sqrt{b}-1\right)-\left(\sqrt{b}-1\right)=\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\)
\(=\sqrt{a}\left(\sqrt{a}+1\right)+2\sqrt{b}\left(\sqrt{a}+1\right)=\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\sqrt{b}\right)\)
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Lời giải:
$x-5\sqrt{x}+6=x-2\sqrt{x}-3\sqrt{x}+6$
$=\sqrt{x}(\sqrt{x}-2)-3(\sqrt{x}-2)$
$=(\sqrt{x}-2)(\sqrt{x}-3)$
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\(x-6\sqrt{x}+8\)
\(=x-2\sqrt{x}-4\sqrt{x}+8\)
\(=\sqrt{x}\left(\sqrt{x}-2\right)-4\left(\sqrt{x}-2\right)\)
\(=\left(\sqrt{x}-2\right)\left(\sqrt{x}-4\right)\)