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\(x+7\sqrt{x}+10=\left(\sqrt{x}+2\right)\left(\sqrt{x}+5\right)\)
\(x-\sqrt{x}-6=\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)\)
\(2x+5\sqrt{x}-3=\left(\sqrt{x}+3\right)\left(2\sqrt{x}-1\right)\)
Lời giải:
$x-5\sqrt{x}+6=x-2\sqrt{x}-3\sqrt{x}+6$
$=\sqrt{x}(\sqrt{x}-2)-3(\sqrt{x}-2)$
$=(\sqrt{x}-2)(\sqrt{x}-3)$
\(a\sqrt{a}+2a+\sqrt{a}+2=\left(a\sqrt{a}+2a\right)+\left(\sqrt{a}+2\right)\)
\(=a\left(\sqrt{a}+2\right)+\left(\sqrt{a}+2\right)=\left(\sqrt{a}+2\right)\left(a+1\right)\)
\(2+\sqrt{3}+\sqrt{6}+\sqrt{8}=2+\sqrt{3}+\sqrt{6}+2\sqrt{2}\)
\(=2+\sqrt{3}+\sqrt{2}\left(2+\sqrt{3}\right)=\left(2+\sqrt{3}\right)\left(\sqrt{2}+1\right)\)
\(2+\sqrt{3}+\sqrt{6}+\sqrt{8}=\left(\sqrt{2}+1\right)\left(2+\sqrt{3}\right)\)
Bài làm:
Ta có: \(-6x+5\sqrt{x}+1\)
\(=\left(-6x+6\sqrt{x}\right)-\left(\sqrt{x}-1\right)\)
\(=-6\sqrt{x}\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)\)
\(=\left(-6\sqrt{x}-1\right)\left(\sqrt{x}-1\right)\)
\(=\left(6\sqrt{x}+1\right)\left(1-\sqrt{x}\right)\)
\(=\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\)
\(5\sqrt{14}-\sqrt{21}=\sqrt{7}\left(5\sqrt{2}-\sqrt{3}\right)\)