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\(12-\sqrt{x}-x\)
ĐK : x ≥ 0
\(=12-4\sqrt{x}+3\sqrt{x}-x\)
\(=4\left(3-\sqrt{x}\right)+\sqrt{x}\left(3-\sqrt{x}\right)\)
\(=\left(3-\sqrt{x}\right)\left(4+\sqrt{x}\right)\)
\(12-\sqrt{x}-x=\left(12-4\sqrt{x}\right)+\left(3\sqrt{x}-x\right)\)
\(=4\left(3-\sqrt{x}\right)+\sqrt{x}\left(3-\sqrt{x}\right)\)
\(=\left(4+\sqrt{x}\right)\left(3-\sqrt{x}\right)\)

\(x\sqrt{x}+4x-12\sqrt{x}-27\)
\(=\left(x\sqrt{x}-27\right)+\left(4x-12\sqrt{x}\right)\)
\(=\left(\sqrt{x}-3\right)\left(x+3\sqrt{x}+9\right)+4\sqrt{x}\left(\sqrt{x}-3\right)\)
\(=\left(\sqrt{x}-3\right)\left(x+3\sqrt{x}+9+4\sqrt{x}\right)\)
\(=\left(\sqrt{x}-3\right)\left(x+7\sqrt{x}+9\right)\)
a, \(\sqrt{a^2-b^2}-\sqrt{a^3+b^3}\)
\(=\sqrt{\left(a+b\right)\left(a-b\right)}-\sqrt{\left(a+b\right)\left(a^2-ab+b^2\right)}\)
\(=\sqrt{a+b}\left(\sqrt{a-b}-\sqrt{a^2-ab+b^2}\right)\)

\(xy-y\sqrt{x}+\sqrt{x}-1\)
\(=y\left(x-\sqrt{x}\right)+\left(\sqrt{x}-1\right)\)
\(=y\sqrt{x}\left(\sqrt{x}-1\right)+\left(\sqrt{x}-1\right)\)
\(\left(\sqrt{x}-1\right)\left(y\sqrt{x}+1\right)\)


\(a)\) \(xy-y\sqrt{x}+\sqrt{x}-1\)
= \(y\sqrt{x}.(\sqrt{x}-1)+\sqrt{x}-1\)
=\((\sqrt{x}-1).(y\sqrt{x}+1)\).
\(b)\)\(\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{ay}\)
=\(\sqrt{a}.\sqrt{x}-\sqrt{b}.\sqrt{y}+\sqrt{b}.\sqrt{x}-\sqrt{a}.\sqrt{y}\)
=\(\sqrt{a}.\sqrt{x}+\sqrt{b}.\sqrt{x}-\sqrt{a}.\sqrt{y}-\sqrt{b}.\sqrt{y}\)
=\(\sqrt{x}.(\sqrt{a}+\sqrt{b})-\sqrt{y}.(\sqrt{a}+\sqrt{b})\)
=\((\sqrt{x}-\sqrt{y}).(\sqrt{a}+\sqrt{b})\).
\(c)\)\(\sqrt{a+b}+\sqrt{a^2-b^2}\)
=\(\sqrt{a+b}+\sqrt{(a+b).(a-b)}\)
=\(\sqrt{a+b}+\sqrt{a+b}.\sqrt{a-b}\)
=\(\sqrt{a+b}.\left(1+\sqrt{a-b}\right)\).
\(d)\) \(12-\sqrt{x}-x\)
=\(12-4\sqrt{x}+3\sqrt{x}-x\)
=\(4.\left(3-\sqrt{x}\right)+\sqrt{x}\left(3-\sqrt{x}\right)\)
=\(\left(3-\sqrt{x}\right).\left(4+\sqrt{3}\right)\).
làm mà chả cho cái đúng nào làm chi
\(12-\sqrt{x}-x\)
\(=12+3\sqrt{x}-4\sqrt{x}-x\)
\(=3.\left(4+\sqrt{x}\right)-\sqrt{x}\left(4+\sqrt{x}\right)\)
\(=\left(4+\sqrt{x}\right)\left(3-\sqrt{x}\right)\)